Answer
$$ \lim _{h \rightarrow 0} \frac{\sqrt[4]{1+h}-1}{h}=\frac{1}{4}$$
Work Step by Step
Given $$ \lim _{h \rightarrow 0} \frac{\sqrt[4]{1+h}-1}{h}$$
Let $$ f(h) = \frac{\sqrt[4]{1+h}-1}{h}$$
Since, we have
$$ f( 0= \frac{1-1}{0}=\frac{0}{0}$$
Let $\sqrt[4]{1+h}=x \ \rightarrow h=x^4-1$
Also, when $ h\rightarrow 0, \ \ \Rightarrow x \rightarrow 1 $,
So, transform algebraically and cancel:
\begin{aligned}
L&= \lim _{h \rightarrow 0} \frac{\sqrt[4]{1+h}-1}{h}\\
& = \lim _{x \rightarrow 1} \frac{x-1}{x^{4}-1}\\
& = \lim _{x \rightarrow 1} \frac{x-1}{(x^{2}-1)(x^2+1)}\\
&=\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(x+1)\left(x^{2}+1\right)}\\
&=\lim _{x \rightarrow 1} \frac{1}{(x+1)\left(x^{2}+1\right)}\\
&=\frac{1}{(1+1)\left(1^{2}+1\right)}\\
&=\frac{1}{4}
\end{aligned}
