# JEE Main & Advanced Mathematics Functions Methods of Evaluation of Limits

## Methods of Evaluation of Limits

Category : JEE Main & Advanced

We shall divide the problems of evaluation of limits in five categories.

(1) Algebraic limits : Let $f(x)$ be an algebraic function and $'a'$ be a real number. Then $\underset{x\to a}{\mathop{\lim }}\,f(x)$ is known as an algebraic limit.

(i) Direct substitution method : If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression.

(ii) Factorisation method : In this method, numerator and denominator are factorised. The common factors are cancelled and the rest outputs the results.

(iii) Rationalisation method : Rationalisation is followed when we have fractional powers (like $\frac{1}{2},\frac{1}{3}$ etc.) on expressions in numerator or denominator or in both. After rationalisation the terms are factorised which on cancellation gives the result.

(iv) Based on the form when $x\to \infty$ : In this case expression should be expressed as a function $1/x$ and then after removing indeterminate form, (if it is there) replace $\frac{1}{x}$ by 0.

(2) Trigonometric limits : To evaluate trigonometric limit the following results are very important.

(i) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x}{\sin x}$

(ii) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\tan x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x}{\tan x}$

(iii) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{\sin }^{-1}}x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\sin }^{-1}}x}$

(iv) $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\tan }^{-1}}x}{x}=1=\underset{x\to 0}{\mathop{\lim }}\,\frac{x}{{{\tan }^{-1}}x}$

(v) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sin {{x}^{0}}}{x}=\frac{\pi }{180}$

(vi) $\underset{x\to 0}{\mathop{\lim }}\,\,\cos x=1$

(vii) $\underset{x\to a}{\mathop{\lim }}\,\,\frac{\sin (x-a)}{x-a}=1$

(viii) $\underset{x\to a}{\mathop{\lim }}\,\,\frac{\tan (x-a)}{x-a}=1$

(ix) $\underset{x\to a}{\mathop{\lim }}\,{{\sin }^{-1}}x={{\sin }^{-1}}a,\,\,|a|\,\,\le 1$

(x) $\underset{x\to a}{\mathop{\lim }}\,\,{{\cos }^{-1}}\,x={{\cos }^{-1}}\,a;\,\,|a|\,\,\le 1$

(xi) $\underset{x\to a}{\mathop{\lim }}\,\,{{\tan }^{-1}}\,x={{\tan }^{-1}}a;\,\,-\infty <a<\infty$

(xii) $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin x}{x}=\underset{x\to \infty }{\mathop{\lim }}\,\frac{\cos x}{x}=0$

(xiii) $\underset{x\to \infty }{\mathop{\lim }}\,\frac{\sin \left( 1/x \right)}{\left( 1/x \right)}=1$

(3) Logarithmic limits : To evaluate the logarithmic limits we use following formulae

(i) $\log (1+x)=x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-............\text{to}\,\infty$ where $-1<x\le 1$ and expansion is true only if base is e.

(ii) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\log (1+x)}{x}=1$

(iii) $\underset{x\to e}{\mathop{\lim }}\,\,{{\log }_{e}}x=1$

(iv) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\log (1-x)}{x}=-1$

(v) $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{a}}(1+x)}{x}={{\log }_{a}}e,\,a>0,\ne 1$

(4) Exponential limits

(i) Based on series expansion

We use ${{e}^{x}}=1+x+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{3}}}{3\,!}+.............\infty$

To evaluate the exponential limits we use the following results

(a) $\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{x}}-1}{x}=1$

(b) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{a}^{x}}-1}{x}={{\log }_{e}}a$

(c) $\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\lambda x}}-1}{x}=\,\lambda \,\,(\lambda \ne 0)$

(ii) Based on the form ${{1}^{\infty }}$ :  To evaluate the exponential form ${{1}^{\infty }}$ we use the following results.

(a) If $\underset{x\to a}{\mathop{\lim }}\,\,f(x)=\underset{x\to a}{\mathop{\lim }}\,\,g(x)=0$, then

$\underset{x\to a}{\mathop{\lim }}\,\,{{\{1+f(x)\}}^{1/g(x)}}\,\,=\,{{e}^{\underset{x\to a}{\mathop{\lim }}\,\,\frac{f(x)}{g(x)}}}$  or when $\underset{x\to a}{\mathop{\lim }}\,\,f(x)=1$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)=\infty$.

Then $\underset{x\to a}{\mathop{\lim }}\,{{\{f(x)\}}^{g(x)}}=\underset{x\to a}{\mathop{\lim }}\,\,{{[1+f(x)-1]}^{g(x)}}$$={{e}^{\underset{x\to a}{\mathop{\lim }}\,(f(x)-1)g(x)}}$

(b) $\underset{x\to 0}{\mathop{\lim }}\,{{(1+x)}^{1/x}}=e$

(c) $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{1}{x} \right)}^{x}}=e$

(d) $\underset{x\to 0}{\mathop{\lim }}\,{{(1+\lambda x)}^{1/x}}={{e}^{\lambda }}$

(e) $\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{\lambda }{x} \right)}^{x}}={{e}^{\lambda }}$

• $\underset{x\to \infty }{\mathop{\lim }}\,\,{{a}^{x}}=\left\{ \begin{matrix} \infty\,,\,\text{if}\,\,a>1 \\ 0\,\,,\text{if}\,\,a \end{matrix}\right.$ e., ${{a}^{\infty }}=\infty$, if $a>1$ and ${{a}^{\infty }}=0$ if $a<1$.

(5) L-Hospital’s rule : If $f(x)$ and $g(x)$ be two functions of $x$ such that

(i) $\underset{x\to a}{\mathop{\lim }}\,f(x)=\underset{x\to a}{\mathop{\lim }}\,g(x)=0$

(ii) Both are continuous at $x=a$

(iii) Both are differentiable at $x=a$.

(iv) $f'(x)$ and $g'(x)$ are continuous at the point $x=a$, then $\underset{x\to a}{\mathop{\lim }}\,\frac{f(x)}{g(x)}\,=\,\underset{x\to a}{\mathop{\lim }}\,\frac{f'(x)}{g'(x)}$ provided that $g'(a)\ne 0$.

The above rule is also applicable if $\underset{x\to a}{\mathop{\lim }}\,\,f(x)=\infty$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)=\infty$.

If $\underset{x\to a}{\mathop{\lim }}\,\,\frac{f'(x)}{g'(x)}$ assumes the indeterminate form $\tfrac{0}{0}$ or $\frac{\infty }{\infty }$ and $f'(x),g'(x)$ satisfy all the condition embodied in L’ Hospital rule, we can repeat the application of this rule on $\frac{f'(x)}{g'(x)}$ to get, $\underset{x\to a}{\mathop{\lim }}\,\,\frac{f'(x)}{g'(x)}=\underset{x\to a}{\mathop{\lim }}\,\frac{f''(x)}{g''(x)}$. Sometimes it may be necessary to repeat this process a number of times till our goal of evaluating limit is achieved.

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