Limits with two unknowns (x,h)

Question

I have a question where there are two unknows ($x$ and $h$) How should I approach the question?

$$\lim\limits_{h\to 0}\frac{\dfrac{1}{\sqrt[3]{x+h}} - \dfrac{1}{\sqrt[3]{x}}}{h}\tag{1}$$

$$\lim\limits_{h\to 0}\frac{\dfrac{x-h}{(x-h-3)^2} - \dfrac{x}{(x-3)^2}}{h}\tag{2}$$

Answer 1

Recall the definition of a derivative:

$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(h)}{h}$$

So the basis of your question is asking for the derivative of

$$f(x)=\frac{1}{\sqrt[3]{x}}$$ and $$g(x)=\frac{x}{(x-3)^3}$$

You could simply apply the formulas and theorems regarding both limits, but here is a more rigprous approach of the first

$$f'(x)=\lim_{h\rightarrow 0}\frac{\frac{1}{\sqrt[3]{x+h}}-\frac{1}{\sqrt[3]{x}}}{h}\\ \lim_{h\rightarrow 0}\frac{\frac{1}{\sqrt[3]{x+h}}-\frac{1}{\sqrt[3]{x}}}{h}=\lim_{h\rightarrow 0}\frac{\frac{\sqrt[3]{x}-\sqrt[3]{x+h}}{\sqrt[3]{x+h}\sqrt[3]{x}}}{h}\\ \lim_{h\to 0}\frac{\sqrt[3]{x}-\sqrt[3]{x+h}}{h\sqrt[3]{(x+h)x}}$$

Recall $x^3-y^3=(x-y)(x^2+xy+y^2)$

$$\lim_{h\to 0}\frac{\sqrt[3]{x}-\sqrt[3]{x+h}}{h\sqrt[3]{(x+h)x}}\frac{\sqrt[3]{x^2}+\sqrt[3]{(x+h)x}+\sqrt[3]{(x+h)^2}}{\sqrt[3]{x^2}+\sqrt[3]{(x+h)x}+\sqrt[3]{(x+h)^2}}=\lim_{h\to 0}{\frac{x-(x+h)}{(h\sqrt[3]{(x+h)x})(\sqrt[3]{x^2}+\sqrt[3]{(x+h)x}+\sqrt[3]{(x+h)^2})}}\\ \lim_{h\to 0}\frac{-h}{{(h\sqrt[3]{(x+h)x})(\sqrt[3]{x^2}+\sqrt[3]{(x+h)x}+\sqrt[3]{(x+h)^2})}}=\frac{-1}{(\sqrt[3]{x^2})(\sqrt[3]{x^2}+\sqrt[3]{x^2}+\sqrt[3]{x^2})}\\ \frac{-1}{\sqrt[3]{x^2}(3\sqrt[3]{x^2})}=\boxed{\frac{-1}{3x\sqrt[3]{x}}}$$

Answer 2

In each of these limits, you regard $x$ as a fixed constant and let $h\to 0$.