GATE CS & DA Preparation Guide Module 4 – Engineering mathematics (CS section 1; DA sections 1-3)

Limits, MVT, integrals, Taylor series and optima

GATE CS and DA calculus: limits, continuity and differentiability, the mean value theorem, definite integrals, Taylor series and one-variable optima.

Intermediate
20 minutes
Examples run with Python 3.14.8, sympy 1.14.0, scipy 1.18.1 and Pyodide 314.0.7
By MySmartCoPilot

Free: Limits: try algebra and series before the derivative rule · Continuity needs the values to meet, differentiability the slopes too · Definite integrals: look for symmetry before computing · 2 more sections. With a pass: 2 in-depth parts and 3 interview answers, about 12 minutes.

What you will learn

Evaluate limits by algebra, standard limits and series, using L'Hôpital's rule only on 0/0 and ∞/∞ forms
Test continuity and differentiability at a point, including piecewise functions
Apply the mean value theorem and evaluate definite integrals with standard techniques
Use Taylor series for limits and approximations, with a bound on the error
Find the global optima of a function on a closed interval by comparing every candidate

Before you start

MCQ, MSQ and NAT: GATE marking and when to guess

What you’ll learn · Before you start

What you will learn

Evaluate limits by algebra, standard limits and series, using L'Hôpital's rule only on 0/0 and ∞/∞ forms
Test continuity and differentiability at a point, including piecewise functions
Apply the mean value theorem and evaluate definite integrals with standard techniques
Use Taylor series for limits and approximations, with a bound on the error
Find the global optima of a function on a closed interval by comparing every candidate

Before you start

MCQ, MSQ and NAT: GATE marking and when to guess

On this page

(1 − cos x)/x² tends to 1/2 as x → 0: a NAT question wants one such number, which a fixed method finds. Section 1 of the GATE CS syllabus lists calculus as limits, continuity and differentiability, maxima and minima, the mean value theorem and integration; section 3 of the DA syllabus adds Taylor series and one-variable optimisation.

Limits yield to algebra, standard limits, series or L’Hôpital’s rule, which applies only to 0/0 and ∞/∞ forms. By the mean value theorem (MVT), a function continuous on [a, b] and differentiable on (a, b) has a point c in (a, b) where f′(c) equals the slope of the chord. Definite integrals rest on substitution, integration by parts and a few symmetry properties. Taylor series give limits and approximations, with a bound on the error. Global optima on [a, b] come from comparing f at the endpoints and the critical points.

Not affiliated

MySmartCoPilot is independent of IIT Madras, IISc and the IITs.

Limits: try algebra and series before the derivative rule

1. Limits (NAT). (1 − cos x)/x² → 1/2 as x → 0, because 1 − cos x = x²/2 − x⁴/24 + ⋯.

Substitute first. If that gives 0/0 or ∞/∞, an indeterminate form: A limit whose shape, such as 0/0 or ∞/∞, does not decide its value. It needs algebra, a series or L'Hôpital's rule, which applies directly to these two shapes only., factor, use standard limits or series, and only then L’Hôpital’s rule. The trap is L’Hôpital on a form that is not indeterminate. Step through the tools on example 1:

Interactive · stage by stage Which tool for a limit: (1 − cos x)/x² as x → 0 Step through the tools in the order to try them. Each step says what it does to this limit.

The drawing shows the whole scene; its 5 steps are listed below.

Every step as text

Substitute first. Put x = 0 into (1 − cos x)/x². The top is 1 − 1 = 0 and the bottom is 0, so substitution gives 0/0, which is not a number.
Name the form. 0/0 is an indeterminate form, so its shape says nothing about the value. Only 0/0 and ∞/∞ qualify for L'Hôpital's rule later on.
Algebra. Factoring, cancelling or rationalising removes many 0/0 forms. Here 1 − cos x has no visible factor x to cancel, so move on.
A series. The series of cos x gives 1 − cos x = x²/2 − x⁴/24 + ⋯, so the ratio is 1/2 − x²/24 + ⋯, which tends to 1/2.
The rule, last. L'Hôpital's rule works here too, twice, through sin x/(2x) and cos x/2, and gives 1/2. Keep it for genuine 0/0 or ∞/∞ forms, after the other tools.

Checkpoint

Check: Limits by algebra, series and the derivative rule

Question 1 of 2 What is the limit of (x² − 9)/(x − 3) as x → 3?
Choose one answer.

It does not exist, because the bottom is 0 at x = 3 0, because the top is 0 at x = 3 ∞, because the bottom tends to 0 6, after factoring x² − 9 and cancelling x − 3

Show the answer: Question 1 of 2

Answer: 6, after factoring x² − 9 and cancelling x − 3

Substitution gives 0/0, an indeterminate form, so simplify first: x² − 9 = (x − 3)(x + 3), and the ratio is x + 3 for every x ≠ 3, which tends to 6.
Question 2 of 2 A candidate applies L'Hôpital's rule to (x + 4)/(2x + 1) as x → 0 and gets 1/2. What is the actual limit?
Type a number.
Your answer

A hint

Substitute x = 0 first, before reaching for any rule.

Show the answer: Question 2 of 2

Answer: 4

Substitution gives 4/1 = 4, a number, so the limit is 4. The form is not 0/0 or ∞/∞, so the rule does not apply, and differentiating top and bottom gave the wrong 1/2.

Four series turn a limit into arithmetic

Four Maclaurin series: The Taylor series of a function about 0: the sum of f⁽ⁿ⁾(0)xⁿ/n! for n = 0, 1, 2, …, such as 1 + x + x²/2! + x³/3! + ⋯ for eˣ., Taylor series about 0, do most of the work:

Function	Its Maclaurin series
eˣ	1 + x + x²/2! + x³/3! + ⋯
sin x	x − x³/3! + x⁵/5! − ⋯
cos x	1 − x²/2! + x⁴/4! − ⋯
ln(1 + x)	x − x²/2 + x³/3 − ⋯

The trap is stopping before the first term that survives.

Stopping one term too early

In (e^(2x) − 1 − 2x)/x², stopping at e^(2x) ≈ 1 + 2x leaves 0 on top; the next term, 2x², survives, and the limit is 2, as the SymPy run below confirms.

Checkpoint

Check: Limits by algebra, series and the derivative rule

Question 1 of 2 How does the Maclaurin series of sin x begin?
Choose one answer.

x − x³/6 + x⁵/120 − ⋯ x − x²/2 + x³/3 − ⋯ 1 − x²/2 + x⁴/24 − ⋯ x + x³/6 + x⁵/120 + ⋯

Show the answer: Question 1 of 2

Answer: x − x³/6 + x⁵/120 − ⋯

sin x = x − x³/3! + x⁵/5! − ⋯ = x − x³/6 + x⁵/120 − ⋯, the row of the table. Only odd powers appear, because sin is an odd function.
Question 2 of 2 Evaluate the limit of (ln(1 + x) − x)/x² as x → 0.
Type a number.
Your answer

A hint

Write ln(1 + x) from the table, up to the x² term.

Show the answer: Question 2 of 2

Answer: -0.5

ln(1 + x) − x = −x²/2 + x³/3 − ⋯, so the ratio is −1/2 + x/3 − ⋯, which tends to −1/2. Stopping at ln(1 + x) ≈ x leaves 0 on top and no answer.

Continuity needs the values to meet, differentiability the slopes too

2. Differentiability (MCQ). |x − 2| is continuous at 2 but not differentiable there: the derivative is −1 from the left and +1 from the right.

To test a point, compare the left and right limits with the value, then the left and right one-sided derivative: The limit of (f(a + h) − f(a))/h as h tends to 0 from one side only. f is differentiable at a exactly when both one-sided derivatives exist and are equal.. Differentiability at a point implies continuity there, but not the other way round. The trap: |x| is continuous at 0 but not differentiable there. Step through both tests on example 2:

Interactive · stage by stage |x − 2| at 2: continuous, but not differentiable Step through the two tests in order. The values must meet for continuity, and the slopes as well for differentiability.

The drawing shows the whole scene; its 5 steps are listed below.

Every step as text

The value. At x = 2 the function is |2 − 2| = 0.
The limits. Just left of 2, f(x) = 2 − x, and just right of it, f(x) = x − 2. Both tend to 0 as x tends to 2.
Continuity. The two one-sided limits equal f(2), so f is continuous at 2. The graph has no jump there.
The slopes. The one-sided derivatives are the slopes of the two pieces, −1 from the left and +1 from the right.
Differentiability. The slopes differ, so f is not differentiable at 2. The graph has a corner and no tangent there, although it is continuous.

Checkpoint

Check: Continuity, differentiability and the MVT

Question 1 of 2 Which is true of f(x) = |x + 1| at x = −1?
Choose one answer.

It is continuous but not differentiable It is differentiable but not continuous It is differentiable, with derivative 0 It is neither continuous nor differentiable

Show the answer: Question 1 of 2

Answer: It is continuous but not differentiable

|x + 1| is |x − 2| moved along: values that meet at −1 and slopes −1 and +1 that do not, so it is continuous and not differentiable there.
Question 2 of 2 f(x) = x + 1 for x ≤ 2 and f(x) = kx for x > 2. For which k is f continuous at 2?
Type a number.
Your answer

A hint

The two pieces must take the same value at x = 2.

Show the answer: Question 2 of 2

Answer: 1.5

The left piece gives f(2) = 3 and the right piece tends to 2k, so 2k = 3 and k = 1.5. f is still not differentiable at 2, because the slopes 1 and 1.5 differ.

The mean value theorem promises a tangent parallel to the chord

3. The MVT (NAT). For f(x) = x² on [1, 3] the chord slope is (9 − 1)/2 = 4, and f′(c) = 2c = 4 gives c = 2.

Rolle’s theorem is the case f(a) = f(b), where f′(c) = 0. Check the conditions, then solve f′(c) = (f(b) − f(a))/(b − a). The trap is a c outside (a, b), or a corner inside it. Step through example 3:

Interactive · stage by stage The mean value theorem for x² on [1, 3] Step through the theorem. Check its conditions, take the chord's slope, then find where the tangent matches it.

The drawing shows the whole scene; its 4 steps are listed below.

Every step as text

The conditions. x² is continuous on [1, 3] and differentiable on (1, 3), as every polynomial is, so the theorem applies.
The chord. The chord joins (1, 1) and (3, 9), so its slope is (9 − 1)/(3 − 1) = 4.
The tangent. The theorem promises a c in (1, 3) where the tangent is parallel to the chord, so f′(c) = 2c must equal 4.
The point. c = 2, inside (1, 3). A solution outside the open interval would be a slip, because the theorem promises one inside.

Checkpoint

Check: Continuity, differentiability and the MVT

Question 1 of 2 Find the value c of the mean value theorem for f(x) = 1/x on [1, 4].
Type a number.
Your answer

A hint

The chord's slope is (f(4) − f(1))/(4 − 1).

Show the answer: Question 1 of 2

Answer: 2

1/x is continuous on [1, 4] and differentiable on (1, 4). The chord slope is (1/4 − 1)/3 = −1/4, and f′(c) = −1/c² = −1/4 gives c² = 4, so c = 2, inside (1, 4).
Question 2 of 2 f(x) = 1/x² has f(−1) = f(1) = 1, yet no c in (−1, 1) has f′(c) = 0. Which condition of Rolle's theorem fails?
Choose one answer.

Equal values at the ends None, since c = 0 works Continuity on [−1, 1], because f is not defined at 0 Differentiability at the endpoints

Show the answer: Question 2 of 2

Answer: Continuity on [−1, 1], because f is not defined at 0

Rolle's theorem needs continuity on [a, b], differentiability on (a, b) and f(a) = f(b). The middle of [−1, 1] holds a point where f is undefined, so the first condition fails.

Definite integrals: look for symmetry before computing

4. Integrals (NAT). On [−1, 1], x³ is odd and x² even, so the integral of x³ + x² is 0 + 2/3 ≈ 0.667.

The tools are substitution, parts, odd and even symmetry, and f(x) against f(a − x). Integrals of an odd function: A function with f(−x) = −f(x), such as x³ or sin x. Its integral over any interval [−a, a] is 0. over [−a, a] are 0, and of an even function twice the integral over [0, a]. The trap is keeping the old limits after a substitution. The program checks five answers against the values worked by hand:

Five limits and integrals, a symmetric integral and two derivatives Python · calc_check.py

import math

from scipy.integrate import quad
from sympy import Abs, E, Rational, diff, exp, integrate, limit, oo, sin, symbols

x = symbols("x", real=True)

checks = [
    ("lim x->0 (e^(2x) - 1 - 2x) / x^2", limit((exp(2 * x) - 1 - 2 * x) / x**2, x, 0), 2),
    ("lim x->0 (sin 3x - 3x) / x^3", limit((sin(3 * x) - 3 * x) / x**3, x, 0), Rational(-9, 2)),
    ("lim x->oo (1 + 3/x)^(2x)", limit((1 + 3 / x) ** (2 * x), x, oo), E**6),
    ("integral of x e^x from 0 to 1", integrate(x * exp(x), (x, 0, 1)), 1),
    ("integral of |x - 1| from 0 to 3", integrate(Abs(x - 1), (x, 0, 3)), Rational(5, 2)),
]
for label, sympy_value, by_hand in checks:
    agree = sympy_value == by_hand
    print(f"{label:<34} SymPy {str(sympy_value):<7} by hand {str(by_hand):<7} agree {agree}")

# With f(x) + f(pi/2 - x) = 1 on [0, pi/2], the integral of f is half the length of the interval: pi/4.
value, error = quad(lambda t: math.sin(t) ** 3 / (math.sin(t) ** 3 + math.cos(t) ** 3), 0, math.pi / 2)
print(f"integral of sin^3 / (sin^3 + cos^3) on [0, pi/2]: quad {value:.12f}, pi/4 = {math.pi / 4:.12f}")

# One-sided difference quotients at 0: x|x| is differentiable there, |x| is not.
h = symbols("h", positive=True)
for f in (x * Abs(x), Abs(x)):
    right = limit((f.subs(x, h) - f.subs(x, 0)) / h, h, 0)
    left = limit((f.subs(x, -h) - f.subs(x, 0)) / (-h), h, 0)
    verdict = "differentiable" if left == right else "not differentiable"
    print(f"f(x) = {f}: derivative from the left {left}, from the right {right}: {verdict} at 0")
xp, xn = symbols("x", positive=True), symbols("x", negative=True)
print(f"derivative of x|x|: {diff(xp * Abs(xp), xp)} for x > 0 and {diff(xn * Abs(xn), xn)} for x < 0, so 2|x| everywhere")

Output

lim x->0 (e^(2x) - 1 - 2x) / x^2   SymPy 2       by hand 2       agree True
lim x->0 (sin 3x - 3x) / x^3       SymPy -9/2    by hand -9/2    agree True
lim x->oo (1 + 3/x)^(2x)           SymPy exp(6)  by hand exp(6)  agree True
integral of x e^x from 0 to 1      SymPy 1       by hand 1       agree True
integral of |x - 1| from 0 to 3    SymPy 5/2     by hand 5/2     agree True
integral of sin^3 / (sin^3 + cos^3) on [0, pi/2]: quad 0.785398163397, pi/4 = 0.785398163397
f(x) = x*Abs(x): derivative from the left 0, from the right 0: differentiable at 0
f(x) = Abs(x): derivative from the left -1, from the right 1: not differentiable at 0
derivative of x|x|: 2*x for x > 0 and -2*x for x < 0, so 2|x| everywhere

Recorded with Python 3.14.8 on macOS 26 arm64. To run it yourself: mise exec python@3.14.8 -- python3 calc_check.py

Each limit and integral agrees with the value worked by hand. The quad line confirms an integral that a symmetry settles in one step: sin³x/(sin³x + cos³x) and its value at π/2 − x add up to 1, so the integral over [0, π/2] is half the length of the interval, π/4. The last lines show the difference between x|x|, which is differentiable at 0 with derivative 2|x|, and |x|, whose derivatives from the left and the right disagree.

Checkpoint

Check: Definite integrals by symmetry

Question 1 of 2 Evaluate the integral of x⁵ + 3 over [−2, 2].
Type a number.
Your answer

Show the answer: Question 1 of 2

Answer: 12

x⁵ is odd, so its integral over [−2, 2] is 0, and the constant 3 gives 3 · 4 = 12.
Question 2 of 2 Evaluate the integral of 2x/(1 + x²) over [1, 2], to three decimals.
Type a number.
Your answer

A hint

Substitute u = 1 + x², and move the limits with it.

Show the answer: Question 2 of 2

Answer: 0.916 (anything from 0.914 to 0.918 counts)

With u = 1 + x², du = 2x dx, and the limits 1 and 2 become 2 and 5, so the integral is ln 5 − ln 2 = ln 2.5 ≈ 0.916. Keeping the old limits would give ln 2 − ln 1 ≈ 0.693.

Global optima on a closed interval sit at endpoints, stationary points or corners

5. Optima (NAT). f(x) = x(6 − x) on [0, 4] has f′(x) = 6 − 2x = 0 at x = 3, f(3) = 9, while f(0) = 0 and f(4) = 8, so the maximum is 9.

The diagram lists where to look:

Where the global optima of f on [a, b] can be

Text description of the diagram

The chart reads from top to bottom.

The goal: the global maximum and minimum of a continuous function f on a closed interval [a, b]. They exist, because f is continuous on a closed and bounded interval.
A group titled "List every candidate" holds three kinds of point: the endpoints a and b; the stationary points, where f'(c) = 0 with c inside (a, b); and the points inside (a, b) where f' does not exist, such as a corner of |x - 1|.
Evaluate f at every candidate.
Compare: the largest value is the global maximum and the smallest is the global minimum.

A point outside [a, b] is never a candidate, and an endpoint is always one.

A continuous function on a closed interval has a global maximum and a global minimum, and each sits at one of three kinds of point. Local optima inside the interval are stationary point: A point where f′(x) = 0, such as x = 3 for 6x − x². With the endpoints and the corners, the stationary points are the candidates for the global optima of f on [a, b]. or corners, so the endpoints are the only extra candidates. Evaluating f at every candidate and comparing is exact, needs no second derivative and cannot miss an endpoint maximum.

The trap is ignoring the endpoints. Step through example 5:

Interactive · stage by stage The closed-interval method on x(6 − x) over [0, 4] Step through the method. List every candidate, evaluate f at each, then compare.

The drawing shows the whole scene; its 4 steps are listed below.

Every step as text

The derivative. f(x) = x(6 − x) = 6x − x², so f′(x) = 6 − 2x, which is 0 at x = 3. A polynomial has no corners, so there are no other points to add.
The candidates. The candidates are the endpoints 0 and 4 and the stationary point 3, which lies inside [0, 4].
The values. Evaluate f at each candidate: f(0) = 0, f(3) = 9 and f(4) = 8.
The verdict. The largest value, 9 at x = 3, is the global maximum, and the smallest, 0 at the endpoint x = 0, is the global minimum. Leaving out the endpoints would lose the minimum.

Checkpoint

Check: Global optima on a closed interval

Question 1 of 2 What is the global maximum of f(x) = x² − 4x on [0, 5]?
Type a number.
Your answer

Show the answer: Question 1 of 2

Answer: 5

f′(x) = 2x − 4 is 0 at x = 2, where f(2) = −4. The endpoints give f(0) = 0 and f(5) = 5, so the maximum is 5, at the endpoint 5; the stationary point is the minimum.
Question 2 of 2 Where on [0, 3] does f(x) = |x − 2| + 1 take its global minimum?
Choose one answer.

At x = 2, a corner where f′ does not exist At the endpoint x = 0 Nowhere, because f′ is never 0 At the endpoint x = 3

Show the answer: Question 2 of 2

Answer: At x = 2, a corner where f′ does not exist

The candidates are the endpoints 0 and 3 and the corner 2. Their values are 3, 2 and 1, so the minimum is 1 at x = 2 and the maximum 3 at x = 0.

Practise: calculus in GATE’s formats

Practise all four ideas, starting with the one you know least.

Taylor series, limits and the MVT: errors measured and conditions tested Pro: opens with a Pro, Premium or Ultimate pass

A series turns a hard limit into arithmetic, and its error bound says how many terms an approximation needs. This part measures that bound, works three limit families by series, shows a limit that L’Hôpital’s rule cannot do, and tests the conditions of Rolle’s theorem and the MVT.

In this part:

Maclaurin polynomials of eˣ at 0.5, with the Lagrange bound measured
Three limit families, worked by series
Piecewise functions: continuity and differentiability, worked
A limit that L’Hôpital’s rule cannot do
Rolle’s theorem and the MVT: conditions tested, c found

2 runnable examples (Python) · about 5 minutes

Opening this part with your pass…

This part opens with any MySmartCoPilot pass.

Pro pass · 1 week₹79$2.99
30 days₹179$5.99
1 year₹1,699$45

One payment; a pass never renews. The price shown is the total you pay: no fees or charges are added at checkout.

Choose a pass I have a licence key

It also opens every other in-depth part in Learn and every Pro tool. Sales are final, except where the law gives you a right to withdraw, for example in the EU and the UK (refund policy).

Integrals and optimisation: properties, improper integrals and the endpoint trap Pro: opens with a Pro, Premium or Ultimate pass

Many definite integrals fall to one property, and the usual optimisation mistake is a candidate left out. This part works the symmetry properties, decides which improper integrals converge, reproduces an optimiser stopping at a local maximum, and ends with an exercise and a ten-minute set.

In this part:

Definite-integral properties, worked
Improper integrals: which converge
The optimiser that stopped at a local maximum, measured
Find the global extrema on a closed interval
- Results of the sample tests
Convexity and the second-derivative test
A ten-minute set, with answers
A rubric for the set

2 runnable examples (Python) · 1 exercise with sample tests · about 6 minutes

Opening this part with your pass…Pro part. It opens with the same pass as the part above.

Interview questions

Fresher · Warm-up

What is the difference between continuity and differentiability at a point, and which one implies the other?

A function is continuous at a when its limit there exists and equals f(a): the graph has no jump or hole. It is differentiable at a when the difference quotient (f(a + h) − f(a))/h has a limit as h → 0, so the graph has a tangent there. Differentiability implies continuity, because f(a + h) − f(a) is the quotient times h, which tends to 0. The reverse fails: |x − 2| is continuous at 2 but has slope −1 from the left and +1 from the right, so it is not differentiable there. To test a glued function, check that the pieces meet, then that their one-sided derivatives agree.
Fresher to mid

Find the global maximum and minimum of f(x) = |x − 1| + |x − 3| on [0, 4], and say where each is reached. Why does setting f′(x) = 0 not work here?

Model answer, Pro. Model answers open with any pass; prices are in the notice on this page. First step and follow-ups
Fresher to mid

A candidate applies L'Hôpital's rule to the limit of (2x + cos x)/x as x → ∞, gets (2 − sin x)/1 and concludes that the limit does not exist. Is the conclusion right? What is the limit, and what exactly does the rule promise?

Model answer, Pro. First step and follow-ups
Fresher to mid

How many terms of the Maclaurin series of eˣ do you need to be sure that the error at x = 0.5 is below 10⁻⁴? Justify the number with an error bound, not with a calculator comparison.

Model answer, Pro. First step and follow-ups

Key takeaways

For a limit, factor, use standard limits or series, and keep L’Hôpital’s rule for genuine 0/0 or ∞/∞ forms.
Differentiable implies continuous, not the reverse; check the MVT’s conditions before solving for c.
For an integral, look for symmetry first, odd or even and f(x) against f(a − x), before computing anything.
For global optima on [a, b], compare f at the endpoints and the critical points, every time.

Terms in this lesson

indeterminate form: A limit whose shape, such as 0/0 or ∞/∞, does not decide its value. It needs algebra, a series or L'Hôpital's rule, which applies directly to these two shapes only.
Maclaurin series: The Taylor series of a function about 0: the sum of f⁽ⁿ⁾(0)xⁿ/n! for n = 0, 1, 2, …, such as 1 + x + x²/2! + x³/3! + ⋯ for eˣ.
odd function: A function with f(−x) = −f(x), such as x³ or sin x. Its integral over any interval [−a, a] is 0.
one-sided derivative: The limit of (f(a + h) − f(a))/h as h tends to 0 from one side only. f is differentiable at a exactly when both one-sided derivatives exist and are equal.
stationary point: A point where f′(x) = 0, such as x = 3 for 6x − x². With the endpoints and the corners, the stationary points are the candidates for the global optima of f on [a, b].

References

GATE 2027 syllabus, Computer Science and Information Technology (section 1) (Indian Institute of Technology Madras (GATE 2027))
GATE 2027 syllabus, Data Science and Artificial Intelligence (section 3) (Indian Institute of Technology Madras (GATE 2027))
MIT OpenCourseWare 18.01SC Single Variable Calculus (Massachusetts Institute of Technology)
SymPy tutorial: calculus (limit, diff, integrate, series) (SymPy Development Team)
scipy.integrate: quad (SciPy)
scipy.optimize.minimize_scalar (SciPy)
math: exp and factorial (Python Software Foundation)

Report a problem with this lesson

In this module: Engineering mathematics (CS section 1; DA sections 1-3)

Logic, sets, lattices and groups for GATE CS Free core, Pro depth: 7 of 20 minutes free
Graphs, counting, recurrences, generating functions Free core, Pro depth: 8 of 20 minutes free
Linear systems, determinants, eigenvalues and LU Free core, Pro depth: 8 of 20 minutes free
Vector spaces, projections, quadratic forms and SVD Free core, Pro depth: 8 of 20 minutes free
Limits, MVT, integrals, Taylor series and optima Free core, Pro depth: 8 of 20 minutes free
Probability and statistics for GATE CS and DA Free core, Pro depth: 8 of 20 minutes free

Your country

Limits, MVT, integrals, Taylor series and optima

What you will learn

Before you start

What you will learn

Before you start

Limits: try algebra and series before the derivative rule

Four series turn a limit into arithmetic

Continuity needs the values to meet, differentiability the slopes too

The mean value theorem promises a tangent parallel to the chord

Definite integrals: look for symmetry before computing

Global optima on a closed interval sit at endpoints, stationary points or corners

Practise: calculus in GATE’s formats

Taylor series, limits and the MVT: errors measured and conditions tested Pro: opens with a Pro, Premium or Ultimate pass

Integrals and optimisation: properties, improper integrals and the endpoint trap Pro: opens with a Pro, Premium or Ultimate pass

Interview questions

What is the difference between continuity and differentiability at a point, and which one implies the other?

Find the global maximum and minimum of f(x) = |x − 1| + |x − 3| on [0, 4], and say where each is reached. Why does setting f′(x) = 0 not work here?

A candidate applies L'Hôpital's rule to the limit of (2x + cos x)/x as x → ∞, gets (2 − sin x)/1 and concludes that the limit does not exist. Is the conclusion right? What is the limit, and what exactly does the rule promise?

How many terms of the Maclaurin series of eˣ do you need to be sure that the error at x = 0.5 is below 10⁻⁴? Justify the number with an error bound, not with a calculator comparison.

Key takeaways

References

Related tools