Prove that 1)Limx--->0loge(1+x)x= 1. 2)Limx-->0ax-1x= logea, a>0,a≠0.
In the second one,its 'a' to the power 'x' minus 1 divided by 'x' to give log 'a' to the base 'e'.
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1 Answers
11Prove it by expansion
Ans 1) \lim_{x\rightarrow 0}\frac{\ln (1+x)}{x}\ =\ \frac{x-\frac{x^{2}}{2!}+ \frac{x^{3}}{3!}-......}{x}\ =\ \lim_{x\rightarrow 0}\ \left(1 - \frac{x}{2!}+\frac{x^{2}}{3!}- ...... \right) = 1
Ans 2)
\lim_{x\rightarrow 0}\frac{a^{x}-1}{x}\ =\ \lim_{x\rightarrow 0}\left(\frac{e^{x\ln a}-1}{x\ln a}(\ln a) \right) =\ ln a \\\\ \\ \left\{bcoz\ \lim_{x\rightarrow 0}\ \frac{e^{x}-1}{x}\ =\ 1 \right\}