Question

Solve the following Questions in the given Attachment​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle1.\rm{\,\,\,\lim_{x\to0}\dfrac{{e}^{x}-1}{\sqrt{1+x}-1}}$

$\displaystyle\rm{=\lim_{x\to0}\dfrac{\left({e}^{x}-1\right)\left(\sqrt{1+x}+1\right)}{\left(\sqrt{1+x}-1\right)\left(\sqrt{1+x}+1\right)}}$

$\displaystyle\rm{=\lim_{x\to0}\dfrac{\left({e}^{x}-1\right)\left(\sqrt{1+x}+1\right)}{\left(\sqrt{1+x}\right)^2-\left(1\right)^2}}$

$\displaystyle\rm{=\lim_{x\to0}\dfrac{\left({e}^{x}-1\right)\left(\sqrt{1+x}+1\right)}{1+x-1}}$

$\displaystyle\rm{=\lim_{x\to0}\dfrac{\left({e}^{x}-1\right)\left(\sqrt{1+x}+1\right)}{x}}$

$\displaystyle\rm{=\lim_{x\to0}\dfrac{{e}^{x}-1}{x}\cdot\lim_{x\to0}\left(\sqrt{1+x}+1\right)}$

$\displaystyle\rm{=1\cdot\left(\sqrt{1+0}+1\right)}$

$\displaystyle\rm{=\left(1+1\right)}$

$\displaystyle\rm{=2}$

$\displaystyle2.\rm{\,\,\,\lim_{x\to\frac{\pi}{2}}\dfrac{cos(x)}{\left(x-\dfrac{\pi}{2}\right)}}$

$\displaystyle=\rm{\lim_{x\to\frac{\pi}{2}}\dfrac{sin\left(\dfrac{\pi}{2}-x\right)}{\left(x-\dfrac{\pi}{2}\right)}}$

$\displaystyle=\rm{-\lim_{x-\frac{\pi}{2}\to0}\dfrac{sin\left(x-\dfrac{\pi}{2}\right)}{\left(x-\dfrac{\pi}{2}\right)}}$

$\displaystyle=\rm{-1}}$