Question

Find
$$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}+(1+x)^{7}-2}{x} .
$$

Answer 1

$\textbf{Answer:}$

$ \frac{15}{2}$

 

$\textbf{Explanation:}$

We are going to make use of L' Hopitals Rule :  \[ \lim_{x \to a}  \frac {f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]

So we have that \[ \lim_{x \to 0} \frac{\sqrt{1+x} +(1+x)^7 -2}{x} \]

\[ = \lim_{x \to 0} \frac{ \frac{1}{2 \sqrt{x+1}} +7(x+1)^6}{1} \]

\[ = \frac{1}{2 \sqrt{0+1}} +7(0+1)^6 \]

\[ = \frac{15}{2} \]