Limit verify that = 0 I am not able to verify this limit, could someone show a step by step solution to this question?
$$
\lim_{h\to 0} \frac{((h+x)+2)^{3/5}-(x+2)^{3/5}}{h}
$$
evaluated at $x=-2$.
I wish to see it derived without using derivatives. 
 A: Without using derivatives you wish to find
$$
\lim_{h \to 0} \frac{((x+h)+2)^{3/5}-(x+2)^{3/5}}{h} \tag{1}
$$
at $x = -2$ then we can evaluate
$$
\lim_{h \to 0} \frac{((-2+h)+2)^{3/5}-(-2+2)^{3/5}}{h} = \lim_{h \to 0} h^{-2/5} \to \infty
$$
Note however that if you wish to find the limit as $x \to -2$ of $(1)$ then you want to consider
$$
\lim_{x \to -2} \lim_{h \to 0} \frac{((x+h)+2)^{3/5}-(x+2)^{3/5}}{h} = \lim_{x \to -2} \frac{3}{5}(x+2)^{-2/5} \to \infty
$$
but you come to the same issue of getting arbitrarily large values as you get closer and closer to your limit point.
A: If you want to evaluate the limit at a particular value of $x$, simply plug in that number and then take the limit.  That is, you get
$$\lim_{h\to0}{((h-2)+2)^{3/5}-(-2+2)\over h}=\lim_{h\to0}{h^{3/5}-0\over h}=\lim_{h\to0}{1\over h^{2/5}}$$
which, being of the form $1/0$, is undefined.
Note, I italicized evaluate to emphasize a conceptual difference between $f'(a)$ and $\lim_{x\to a}f'(x)$.  
A: This limit is equal to finding the derivative for a function:
$$f(x) = (x+2)^\frac{3}{5}$$
At $x = -2$
So, $$f'(x) = \frac{3}{5}*(x+2)^\frac{-2}{5}$$
$$f'(-2) = \frac{3}{5*0} =  \text{undefined}$$
You are incorrect.
