Calculus

Calculus contributes a small but dependable share of the 13-mark Engineering Mathematics block. The syllabus for the CS paper is narrow — limits, continuity and differentiability, mean value theorems, maxima and minima, and integration — and it is examined at a level well below a mathematics degree.

What makes GATE calculus questions non-trivial is not the computation but the hypotheses. Rolle's theorem, the mean value theorem and the extreme value theorem are all short to state and easy to apply. Every one of them fails on a function that violates one condition, and the questions are built on exactly those failures.

So the working method is: state the theorem, then check each hypothesis against the specific interval given. Continuity on a closed interval, differentiability on the open interval, and the endpoint condition are three separate checks, and a question that supplies $|x|$ on $[-1,1]$ is testing the second one specifically.

The second principle is that a limit is about approach, not about the value at the point. A function can have a perfectly good limit at a point where it is undefined, or where its value is something else entirely. Confusing the limit with the value is the single commonest error in the topic.

1. Limits

The limit of $f$ at $a$ exists when the left-hand and right-hand limits both exist and are equal, and it says nothing whatever about $f(a)$.

Three indeterminate forms cover almost every GATE question: $0/0$, $\infty /\infty$ and $\infty -\infty$. The others, such as $1^{\infty }$ and $0^0$, appear occasionally and are handled by taking logarithms.

L'Hopital's rule applies to $0/0$ and $\infty /\infty$ only:

$L={\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

The rule must not be applied to a form that is not indeterminate, which is the standard misuse. If the limit evaluates to $2/0$, applying L'Hopital gives a wrong answer, because the form is infinite rather than indeterminate.

A small set of standard limits removes most computation.

Series expansion is often faster than repeated L'Hopital. Substituting $\sin x=x-x^3/6+\cdots$ and $e^x=1+x+x^2/2+\cdots$ turns a three-step differentiation into one cancellation.

The expansions worth carrying are few, and all are Maclaurin series about the origin.

Keep terms only to the order the denominator requires. For a denominator of $x^3$, expanding the numerator to $x^3$ suffices, and carrying further terms wastes time without changing the answer.

For limits at infinity of rational functions, compare degrees: the limit is 0 if the denominator has higher degree, infinite if the numerator does, and the ratio of leading coefficients if the degrees are equal.

2. Continuity and Differentiability

A function is continuous at $a$ if the limit at $a$ exists, $f(a)$ exists, and they are equal. All three conditions are separate and each can fail alone.

Differentiability implies continuity, but continuity does not imply differentiability. The absolute value function at 0 is the canonical counterexample: continuous everywhere, not differentiable at 0 because the left derivative is $-1$ and the right derivative is $+1$.

Three failure modes are worth recognising on sight.

For a piecewise function, continuity at the join requires equal one-sided limits, and differentiability additionally requires equal one-sided derivatives. GATE questions supply a piecewise function with unknown constants and ask for the values making it differentiable, which means solving both conditions and never just the first.

3. The Mean Value Theorems

Rolle's theorem states that if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there is some $c$ in $(a,b)$ with $f^{\prime }(c)=0$.

The Lagrange mean value theorem drops the endpoint equality:

$f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

Every hypothesis is load-bearing, and questions are built by removing exactly one.

Take $f(x)=|x|$ on $[-1,1]$. It is continuous on the closed interval and satisfies $f(-1)=f(1)=1$, so the endpoint condition holds. But it is not differentiable at 0, which lies inside the open interval, so Rolle's theorem does not apply — and indeed $f^{\prime }$ is never zero.

Take $f(x)=1/x$ on $[-1,1]$. It fails continuity at 0, so no conclusion follows even though the arithmetic of the theorem could be written down.

The theorems assert existence, not uniqueness. There may be several values of $c$, and a question asking "how many values of $c$ satisfy the mean value theorem" is asking you to solve $f^{\prime }(c)=\text{slope}$ and count roots in the open interval.

Cauchy's mean value theorem generalises to two functions and gives $\frac{f^{\prime }(c)}{g^{\prime }(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$, which is the result L'Hopital's rule is proved from.

4. Maxima and Minima

Critical points are where $f^{\prime }(x)=0$ or where $f^{\prime }$ fails to exist. The second condition is routinely forgotten, and it is exactly where a question about $|x|$ or $x^{2/3}$ places its answer.

The second-derivative test classifies a critical point where $f^{\prime }(c)=0$: negative $f^{\prime \prime }(c)$ gives a local maximum, positive gives a local minimum, and zero is inconclusive.

When the second derivative test is inconclusive, use the sign change of $f^{\prime }$, which always works. For $f(x)=x^4$ at 0, $f^{\prime \prime }(0)=0$, but $f^{\prime }$ changes from negative to positive, so it is a minimum. For $f(x)=x^3$ at 0, $f^{\prime }$ does not change sign, so it is neither.

For a global extremum on a closed interval, the extreme value theorem guarantees existence provided $f$ is continuous. The candidates are the critical points and the two endpoints, and evaluating $f$ at all of them and comparing is the complete method.

Omitting the endpoints is the standard error, and it is the difference between a local and a global answer.

The intermediate value theorem is the other existence result worth holding exactly. If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f$ has at least one root in $(a,b)$.

It requires only continuity, not differentiability, and it guarantees at least one root rather than exactly one. Questions asking how many roots a polynomial has in an interval combine it with monotonicity: the theorem gives existence and a constant sign of $f^{\prime }$ gives uniqueness.

The sign of $f^{\prime }$ gives monotonicity and the sign of $f^{\prime \prime }$ gives concavity. A function is increasing where $f^{\prime }>0$ and concave up where $f^{\prime \prime }>0$, and a point where $f^{\prime \prime }$ changes sign is an inflection point.

An inflection point is not an extremum, and a critical point where $f^{\prime \prime }$ vanishes without changing sign is neither, as $x^3$ at the origin shows.

5. Integration

Two substitution patterns and integration by parts cover most of what is asked.

Substitute whenever the integrand contains a function together with a multiple of its own derivative. The clearest signal is a numerator that is exactly the derivative of the denominator, in which case the answer is the logarithm of the denominator.

Integration by parts follows from the product rule:

$\int udv=uv-\int vdu$

Choose $u$ by ILATE: inverse trigonometric, logarithmic, algebraic, trigonometric, exponential, in that priority order.

A small table of standard integrals removes most lookups.

Definite integrals carry properties that shorten work dramatically.

Over a symmetric interval $[-a,a]$, an odd integrand integrates to zero and an even one integrates to twice the half-interval value. Checking parity before integrating is worth a few seconds on every symmetric-interval question.

The king's property states that ${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx$, which converts many trigonometric integrals into a form where the original and the transformed integral add to something trivial.

The fundamental theorem of calculus links the two halves of the subject and supplies a differentiation rule that is asked directly.

If $F(x)={\int }_a^{x}f(t)dt$ with $f$ continuous, then $F^{\prime }(x)=f(x)$. When the upper limit is itself a function, the chain rule applies, so differentiating ${\int }_a^{g(x)}f(t)dt$ gives $f(g(x))g^{\prime }(x)$.

For area, the definite integral gives signed area, so a region partly below the axis requires splitting at the zeros and taking absolute values. Integrating straight through a sign change and calling the result an area is a standard error.

6. Improper Integrals and Convergence

An improper integral has an infinite limit or an unbounded integrand, and the question is whether it converges.

The $p$-test settles most cases. The integral of $1/x^p$ from 1 to infinity converges exactly when $p>1$, and from 0 to 1 it converges exactly when $p<1$. The two conditions are opposite, which is what makes the test worth stating carefully rather than half-remembering.

For a comparison, if $0\le f\le g$ and the integral of $g$ converges, so does that of $f$; if the integral of $f$ diverges, so does that of $g$.

7. Functions of Two Variables

Partial derivatives differentiate with respect to one variable holding the others fixed.

For a smooth function the mixed partials are equal, so $f_{xy}=f_{yx}$, which is Clairaut's theorem and is assumed throughout.

Critical points of $f(x,y)$ satisfy $f_x=f_y=0$, and they are classified by the discriminant

$D=f_{xx}{f}_{yy}-f_{xy}^2$

evaluated at the critical point.

If $D>0$ and $f_{xx}>0$ it is a local minimum; if $D>0$ and $f_{xx}<0$ a local maximum; if $D<0$ it is a saddle point; and if $D=0$ the test is inconclusive.

The saddle case has no one-variable analogue and is what most two-variable questions are actually testing.

8. Worked Examples

Example 1. Evaluate ${\lim }_{x\to 0}\frac{x-\sin x}{x^3}$.

Substituting gives $0/0$, so the form is indeterminate and L'Hopital applies.

Differentiating once gives $\frac{1-\cos x}{3x^2}$, still $0/0$.

Differentiating again gives $\frac{\sin x}{6x}$, still $0/0$.

Differentiating a third time gives $\frac{\cos x}{6}$, which at 0 is $1/6$.

The series route is faster. Since $\sin x=x-\frac{x^3}{6}+O(x^5)$, the numerator is $\frac{x^3}{6}+O(x^5)$, so the ratio is $\frac{1}{6}+O(x^2)$ and the limit is $1/6$ in one step.

Both agree, and the series method is the one worth defaulting to when three L'Hopital applications would be needed.

Example 2. For what values of $a$ and $b$ is the function $f(x)=x^2$ for $x\le 1$ and $f(x)=ax+b$ for $x>1$ differentiable at $x=1$?

Two conditions must hold, and solving only the first is the standard error.

Continuity at 1 requires the one-sided limits to agree: the left gives $1^2=1$ and the right gives $a+b$. So $a+b=1$.

Differentiability requires the one-sided derivatives to agree: the left derivative is $2x$ at $x=1$, which is 2, and the right derivative is $a$. So $a=2$.

Substituting into the first condition gives $b=-1$.

So $a=2$ and $b=-1$.

Note that with $a=1$ and $b=0$ the function would still be continuous, since $a+b=1$, but the derivative would jump from 2 to 1 and the function would have a corner. Continuity alone never determines both constants.

Example 3. Does Rolle's theorem apply to $f(x)=1-x^{2/3}$ on $[-1,1]$?

Check each hypothesis in turn.

Continuity on the closed interval: $x^{2/3}$ is continuous everywhere, including at 0, so this holds.

Endpoint equality: $f(-1)=1-1=0$ and $f(1)=1-1=0$, so this holds.

Differentiability on the open interval: $f^{\prime }(x)=-\frac{2}{3}{x}^{-1/3}$, which is undefined at $x=0$, a point inside $(-1,1)$. This hypothesis fails.

So Rolle's theorem does not apply.

Confirm the conclusion genuinely fails: $f^{\prime }(x)$ is never zero, since $-\frac{2}{3}{x}^{-1/3}$ has no root. The function does have a maximum at $x=0$, but it is a cusp rather than a stationary point, which is precisely why the differentiability hypothesis was needed.

Example 4. Find the absolute maximum and minimum of $f(x)=x^3-3x+1$ on $[-2,2]$.

The function is a polynomial, so it is continuous on the closed interval and the extreme value theorem guarantees both extrema exist.

Find the critical points: $f^{\prime }(x)=3x^2-3=0$ gives $x=\pm 1$, both inside the interval. There are no points where $f^{\prime }$ fails to exist.

Evaluate at the critical points and both endpoints, which is the complete candidate list.

$f(-2)=-8+6+1=-1$. $f(-1)=-1+3+1=3$. $f(1)=1-3+1=-1$. $f(2)=8-6+1=3$.

The absolute maximum is 3, attained at both $x=-1$ and $x=2$. The absolute minimum is $-1$, attained at both $x=-2$ and $x=1$.

The instructive part is that the endpoint $x=2$ ties the interior maximum. Omitting endpoints would have found the same maximum value here by luck, but would have missed that it is attained twice.

Example 5. Evaluate ${\int }_{-2}^2(x^3+x)dx$.

Check parity before integrating anything.

$f(-x)=-x^3-x=-(x^3+x)=-f(x)$, so the integrand is odd.

The interval $[-2,2]$ is symmetric about the origin, so the integral of an odd function over it is zero.

The answer is 0, with no antiderivative computed at all.

Verify: the antiderivative is $\frac{x^4}{4}+\frac{x^2}{2}$, which is even, so evaluating it at 2 and at $-2$ gives the same value and the difference is zero. The parity check reaches the same place in one line.

Example 6. Classify the critical points of $f(x,y)=x^3-3xy+y^3$.

Set both partials to zero. $f_x=3x^2-3y=0$ gives $y=x^2$, and $f_y=-3x+3y^2=0$ gives $x=y^2$.

Substituting: $x=(x^2{)}^2=x^4$, so $x(x^3-1)=0$, giving $x=0$ or $x=1$.

The critical points are $(0,0)$ and $(1,1)$.

Compute the second partials: $f_{xx}=6x$, $f_{yy}=6y$, $f_{xy}=-3$.

$D=f_{xx}{f}_{yy}-f_{xy}^2=36xy-9$

At $(0,0)$: $D=-9<0$, so it is a saddle point.

At $(1,1)$: $D=36-9=27>0$ and $f_{xx}=6>0$, so it is a local minimum.

The saddle at the origin is the point of the question. Along the line $y=x$ the function is $2x^3-3x^2$, which has a local maximum at the origin, while along $y=-x$ it behaves differently — the two directions disagree, which is exactly what a saddle is.

Summary

The theorems are cheap to state; the marks are in checking their hypotheses against the specific interval given.

A limit describes approach and says nothing about the value at the point. L'Hopital applies only to genuinely indeterminate forms, and series expansion often beats repeated differentiation.

Differentiability implies continuity, never the reverse. Corners, vertical tangents and oscillation are the three failure modes.

For a piecewise function, continuity matches the one-sided limits and differentiability additionally matches the one-sided derivatives; solving only the first never determines both constants.

Rolle's theorem needs continuity on the closed interval, differentiability on the open one, and equal endpoint values; removing any one is how questions are built. The mean value theorems assert existence, not uniqueness.

Critical points include those where the derivative fails to exist. When the second-derivative test is inconclusive, the sign change of $f^{\prime }$ always decides.

For a global extremum on a closed interval, the candidates are the critical points and both endpoints.

Substitute when the integrand contains a function and its derivative; choose $u$ by ILATE for parts. Check parity before integrating over a symmetric interval.

The $p$-test converges for $p>1$ at infinity and for $p<1$ at the origin — opposite conditions.

For two variables, classify critical points by $D=f_{xx}{f}_{yy}-f_{xy}^2$, with $D<0$ giving a saddle, which has no one-variable analogue.