2 votes 2 votes The expression $\lim_{a \to 0}\frac{x^{a}-1}{a}$ is equal to (A)$\log x$ (B)0 (c)$x\log x$ (D)$\infty$ Calculus calculus limits gate-mathematics + – anonymous asked Jul 31, 2016 • edited Apr 9, 2022 by Arjun anonymous 2.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
5 votes 5 votes $\lim_{a \to 0}\frac{x^{a}-1}{a}$ $(\frac{0}{0}\ form )$ 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)$\log x$. LeenSharma answered Jul 31, 2016 LeenSharma comment Share Follow See all 4 Comments See all 4 4 Comments reply anonymous commented Jul 31, 2016 reply Follow Share Thanx. :) 1 votes 1 votes Ritaban Basu_23 commented Sep 13, 2016 reply Follow Share 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. 0 votes 0 votes mcjoshi commented Sep 13, 2016 reply Follow Share In the question above $a\rightarrow0$ not x. So, diffrentiation is done w.r.to a. 1 votes 1 votes Ritaban Basu_23 commented Sep 13, 2016 reply Follow Share Sorry I missed it :) Thanks 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 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, $\LARGE \lim_{a->0}\frac{f(a)-f(0)}{a-0}$ =$\LARGE f'(0)$ =$\LARGE \log x*x^{0}$ =$\LARGE \log x$ Kabir5454 answered May 18, 2022 Kabir5454 comment Share Follow See all 0 reply Please log in or register to add a comment.