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\}