Evaluate $\lim_{h\to 0} \frac{(1+h)^{1/4}-1}{h}$ I have to find the value of  $\frac{(1+h)^{1/4}-1}{h}$ when $h\to 0$
however I can't seem to find the answer. The correct answer is $\frac{1}{4}$ , can someone help me arrive to this answer. 
 A: Use $\LaTeX$ in the future, it helps with readability. 
Now consider that you have: 
$$\lim_{h \to 0} \frac{(1+h)^{1/4}-1}{h} = \lim_{h \to 0} \frac{(1+h)^{1/4}-(1)^{1/4}}{h}$$
In fact, if we let $f(x) = x^{1/4}$, then we have: 
$$\lim_{h \to 0} \frac{(1+h)^{1/4}-1}{h} = \lim_{h \to 0}\frac{f(1+h)-f(1)}{h} = f'(1)$$
That last step is justified by the definition of the derivative. But: 
$$f'(x) = \frac{1}{4}x^{-\frac{3}{4}} \Rightarrow f'(1) = \frac{1}{4}(1)^{-\frac{3}{4}} = \frac{1}{4}$$
Just like we wanted!
A: Set $y^4=1+h$, where $0<|h| <1$,
and consider $\lim y \rightarrow 1 $.
$\dfrac{y-1}{y^4-1}=\dfrac{y-1}{(y-1)(y+1)(y^2+1)}$
$=\dfrac{1}{(y+1)(y^2+1)}$
Take the limit.
A: You can use a binomial series to expand (1+h)^(1/4) (similar example here https://socratic.org/questions/how-do-you-use-the-binomial-series-to-expand-1-4x-1-2). 
Or if we define a function f(x)=(1+x)^1/4 then your limit is actually the definition of the derivative of f(x) at the point x=0. In other words limit of (f(0+h)-f(0))/h as h goes to 0. So alternatively we can compute the derivative of f(x) and substitute x=0 as Anurag A mentioned.
A: That is just a standard limit from the table you should memorize in order to be more flexible:
$$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a$$
Also:
$$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1\;\&\;\lim_{x\to 0}\frac{e^x-1}{x}=1$$
$$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=\lim_{x\to 0}\frac{e^{a\cdot\frac{\ln(1+x)}{x}\cdot x}-1}{ax}\cdot a=a$$
A: There are a few ways you can go.
1) You can say, I recognize this as the definition of the derivative.
$\lim_\limits{h\to 0} \frac {f(x+h) - f(x)}{4}$
With $f(x) = x^\frac 14$ evaluated at $x = 1$
And you really should be looking out for "derivatives in disguise."
2) you can attack it with the binomial theorem
$(a+b)^n= a^n + n a^{n-1}b + \frac {n(n-1)}{2} a^{n-2}b^2 + \cdots$
$a =1, b = h, n = \frac 14$
All of the higher powered terms of $h$ will drop away as $h$ goes to $0$
3) The difference of powers
$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$
How does this help?  Multiply numerator and denominator by $((1+h)^\frac 34 + (1+h)^\frac24 + (1+h)^\frac 14 + 1)$
This will give $\frac {(1+h) - 1}{h((1+h)^\frac 34 + (1+h)^\frac24 + (1+h)^\frac 14 + 1)}$
I think that this is the more elegant algebraically than the binomial theorem. 
The simplify the numerator, and let $h$ go to $0$
A: You can also just rationalize the numerator:
\begin{align*}
\frac{(1+h)^{1/4}-1}{h}
&= \frac{((1+h)^{1/4}-1)((1+h)^{1/4}+1)}{h((1+h)^{1/4}+1)}
= \frac{(1+h)^{1/2}-1}{h((1+h)^{1/4}+1)}\\
&= \frac{((1+h)^{1/2}-1)((1+h)^{1/2}+1)}{h((1+h)^{1/4}+1)((1+h)^{1/2}+1)}
= \frac{h}{h((1+h)^{1/4}+1)((1+h)^{1/2}+1)}\\
&= \frac{1}{((1+h)^{1/4}+1)((1+h)^{1/2}+1)}.
\end{align*}
Then just set $h=0$.
