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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