Gate 2014 CE Set 2

Question

The expression $\lim_{a \to 0}\frac{x^{a}-1}{a}$ is equal to

(A)$\log x$     (B)0          (c)$x\log x$      (D)$\infty$

Answer 1

$\lim_{a \to 0}\frac{x^{a}-1}{a}$                          $(\frac{0}{0}\ form )$

So,applying L-Hospital's rule

$\lim_{a \to 0}\frac{x^{a}\times \log x}{1}$

=$x^{0}\times \log x$

=$\log x$

Hence,Option(A)$\log x$.

Comments

Thanx. :)

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how differentiation of x^a = log x * x^a  
like differentiation of x^2 = 2x

and differentiation of a^x = loga * (a^x)

I think the ans wud be 0.

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In the question above $a\rightarrow0$ not x. So, diffrentiation is done w.r.to a.

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Sorry I missed it :) Thanks

Answer 2

Given,

$\LARGE \lim_{a->0}\frac{x^{a}-1}{a}$          [here x is constant but a is variable ]

Lets take,

$\LARGE f(a)=x^{a}$ ,$\LARGE f(0)=1$

so we can write,

$\LARGE \lim_{a->0}\frac{f(a)-f(0)}{a-0}$

=$\LARGE f'(0)$

=$\LARGE \log x*x^{0}$

=$\LARGE \log x$