Easy Practice — Set 1

Quick Recap — Limits: the Basics

• A limit describes the value $f(x)$ approaches as $x\to a$; it exists iff the left- and right-hand limits are equal.
• Standard limits: $\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1$, $\displaystyle\lim_{x\to0}\frac{\tan x}{x}=1$, $\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\tfrac12$.
• $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$, $\displaystyle\lim_{x\to0}\frac{a^x-1}{x}=\ln a$, $\displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x}=1$.
• $\displaystyle\lim_{x\to a}\frac{x^n-a^n}{x-a}=n\,a^{n-1}$, and $\displaystyle\lim_{x\to\infty}\Big(1+\tfrac1x\Big)^x=e$.

Beyond-NCERT JEE Formulae

L'Hopital's Rule. When $\displaystyle\lim_{x\to a}\dfrac{f(x)}{g(x)}$ has the indeterminate form $\dfrac00$ or $\dfrac{\infty}{\infty}$, the limit equals the ratio of the derivatives:

$$\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)},$$

provided the latter limit exists, and the step may be repeated as long as the indeterminate form persists. [JEE Tip] L'Hopital is a fast alternative to Taylor-series expansion for $\dfrac00$ and $\dfrac{\infty}{\infty}$ limits. [JEE Tip] Differentiate the numerator and the denominator SEPARATELY -- do NOT apply the quotient rule.

Newton-Leibniz Rule. To differentiate an integral whose limits depend on $x$,

$$\dfrac{d}{dx}\int_{g(x)}^{h(x)}f(t)\,dt=f\big(h(x)\big)\,h'(x)-f\big(g(x)\big)\,g'(x).$$

Evaluate the integrand at each limit and multiply by that limit's derivative; the antiderivative of $f$ is never needed.

Handy Limit Forms.

• $\displaystyle\lim_{x\to0}(1+ax)^{1/x}=e^{a}$, which generalises the definition of $e$.
• $\displaystyle\lim_{x\to0}\dfrac{\tan^{-1}x}{x}=1$, and likewise $\displaystyle\lim_{x\to0}\dfrac{\sin^{-1}x}{x}=1$.
• $\displaystyle\lim_{x\to0}\dfrac{a^{x}-b^{x}}{x}=\ln\dfrac{a}{b}$ for $a,b>0$.

Differentiability Shortcuts.

• $|f(x)|$ is NOT differentiable where $f(x)=0$ and $f$ changes sign (a simple root); at a double root, where $f$ keeps its sign, $|f|$ stays smooth.
• $\max\{u,v\}$ and $\min\{u,v\}$ of two differentiable functions have corners only at crossings, where the graphs meet with different slopes.

Solved Examples — Beyond-NCERT Formulae

Each example applies one of the Beyond-NCERT tools above; several are cross-checked by a second method.

Example 1 — A $\dfrac00$ limit by L'Hopital, cross-checked by series

Evaluate $\displaystyle\lim_{x\to0}\dfrac{x-\sin x}{x^{3}}$.

L'Hopital. The form is $\dfrac00$, so differentiate numerator and denominator separately (never as a quotient):

$$\lim_{x\to0}\dfrac{x-\sin x}{x^{3}}=\lim_{x\to0}\dfrac{1-\cos x}{3x^{2}}=\lim_{x\to0}\dfrac{\sin x}{6x}=\dfrac16.$$

Each stage stays $\dfrac00$ until the last, where $\dfrac{\sin x}{6x}\to\dfrac16$.

Series cross-check. From $\sin x=x-\dfrac{x^{3}}{6}+\cdots$, the numerator is $x-\sin x=\dfrac{x^{3}}{6}-\cdots$, so dividing by $x^{3}$ again gives $\dfrac16$. Both routes agree.

Example 2 — An $\dfrac{\infty}{\infty}$ limit by repeated L'Hopital

Evaluate $\displaystyle\lim_{x\to\infty}\dfrac{x^{2}}{e^{x}}$.

The form is $\dfrac{\infty}{\infty}$. Apply L'Hopital and repeat while the form persists:

$$\lim_{x\to\infty}\dfrac{x^{2}}{e^{x}}=\lim_{x\to\infty}\dfrac{2x}{e^{x}}=\lim_{x\to\infty}\dfrac{2}{e^{x}}=0.$$

The exponential outgrows any power of $x$, so the limit is $0$.

Example 3 — Newton-Leibniz with one variable limit

Find $\dfrac{d}{dx}\displaystyle\int_{0}^{x^{2}}\sin t\,dt$.

Here the lower limit is $g(x)=0$ with $g'(x)=0$, the upper limit is $h(x)=x^{2}$ with $h'(x)=2x$, and the integrand is $f(t)=\sin t$. The Newton-Leibniz rule gives

$$\dfrac{d}{dx}\int_{0}^{x^{2}}\sin t\,dt=\sin\!\big(x^{2}\big)\cdot 2x-\sin(0)\cdot 0=2x\sin\!\big(x^{2}\big).$$

Cross-check. Integrating first, $\displaystyle\int_{0}^{x^{2}}\sin t\,dt=\big[-\cos t\big]_{0}^{x^{2}}=1-\cos\!\big(x^{2}\big)$, whose derivative is indeed $2x\sin\!\big(x^{2}\big)$.

Example 4 — Newton-Leibniz with both limits variable

Find $\dfrac{d}{dx}\displaystyle\int_{x}^{x^{3}}e^{t^{2}}\,dt$.

Now the lower limit is $g(x)=x$ with $g'(x)=1$, the upper limit is $h(x)=x^{3}$ with $h'(x)=3x^{2}$, and the integrand is $f(t)=e^{t^{2}}$. The upper limit contributes $f(h)\,h'$ while the lower limit subtracts $f(g)\,g'$:

$$\dfrac{d}{dx}\int_{x}^{x^{3}}e^{t^{2}}\,dt=e^{(x^{3})^{2}}\cdot 3x^{2}-e^{x^{2}}\cdot 1=3x^{2}e^{x^{6}}-e^{x^{2}}.$$

Notice that the antiderivative of $e^{t^{2}}$, which is not elementary, is never required.

Example 5 — A handy exponential form, with a log-and-L'Hopital check

Evaluate $\displaystyle\lim_{x\to0}(1+3x)^{1/x}$.

This matches the standard form $\displaystyle\lim_{x\to0}(1+ax)^{1/x}=e^{a}$ with $a=3$, so the value is $e^{3}$.

Cross-check by logarithm and L'Hopital. Put $L=\displaystyle\lim_{x\to0}(1+3x)^{1/x}$. Taking logs turns the power into a $\dfrac00$ quotient:

$$\ln L=\lim_{x\to0}\dfrac{\ln(1+3x)}{x}=\lim_{x\to0}\dfrac{3/(1+3x)}{1}=3,$$

so $L=e^{3}$, confirming the shortcut.

Example 6 — A $1^{\infty}$ form via logarithm and L'Hopital

Evaluate $\displaystyle\lim_{x\to0}(\cos x)^{1/x^{2}}$.

The base tends to $1$ while the exponent tends to $\infty$, a genuine $1^{\infty}$ form. Set $L=\displaystyle\lim_{x\to0}(\cos x)^{1/x^{2}}$ and take logarithms:

$$\ln L=\lim_{x\to0}\dfrac{\ln\cos x}{x^{2}}.$$

The right side is $\dfrac00$, so by L'Hopital,

$$\ln L=\lim_{x\to0}\dfrac{-\tan x}{2x}=-\dfrac12,$$

using $\dfrac{\tan x}{x}\to1$. Hence $L=e^{-1/2}$.

Example 7 — Handy inverse-trig and difference-of-powers forms

Evaluate $\displaystyle\lim_{x\to0}\dfrac{7^{x}-3^{x}}{x}$ first. Applying $\displaystyle\lim_{x\to0}\dfrac{a^{x}-b^{x}}{x}=\ln\dfrac{a}{b}$ with $a=7$ and $b=3$,

$$\lim_{x\to0}\dfrac{7^{x}-3^{x}}{x}=\ln\dfrac{7}{3}.$$

Next evaluate $\displaystyle\lim_{x\to0}\dfrac{\tan^{-1}5x}{\sin^{-1}2x}$. Using $\dfrac{\tan^{-1}5x}{5x}\to1$ together with $\dfrac{\sin^{-1}2x}{2x}\to1$, the numerator behaves like $5x$ while the denominator behaves like $2x$:

$$\lim_{x\to0}\dfrac{\tan^{-1}5x}{\sin^{-1}2x}=\dfrac{5}{2}.$$