The given limit is a derivative, but of what function and at what point?

Question

How would I solve for something like this??

$$\lim_{h\to 0} \frac{\sqrt[4]{16+h} - 2}{h}$$

using the definition of derivatives.

Answer 1

There are several approaches one can use here to evaluate the limit $L$ given by

$$L= \lim_{h\to 0}\frac{(16+h)^{1/4}-2}{h}$$

METHODOLOGY $1$:

Note that we can rationalize the numerator by multiplying by

$$1=\frac{16+h)^{3/4}+(16+h)^{1/2}(16)^{1/4}+(16+h)^{1/4}(16)^{1/2}+(16)^{3/4}}{(16+h)^{3/4}+(16+h)^{1/2}(16)^{1/4}+(16+h)^{1/4}(16)^{1/2}+(16)^{3/4}}$$

Then, we can have

$$\frac{(16+h)^{1/4}-2}{h}=\frac{1}{(16+h)^{3/4}+(16+h)^{1/2}(16)^{1/4}+(16+h)^{1/4}(16)^{1/2}+(16)^{3/4}}$$

whereupon letting $h\to 0$ reveals

$$\lim_{h\to 0}\frac{(16+h)^{1/4}-2}{h}=\frac{1}{32}$$

METHODOLOGY $2$:

Another approach is to use L'Hospital's Rule, which is tantamount to observing that the limit is the derivative of $f(x)=x^{1/4}$ at $x=16$.

METHODOLOGY $3$:

We can use asymptotic analysis whereby we have

$$\begin{align} (16+h)^{1/4}-2&=2\left(1+\frac h{16}\right)^{1/4}-2\\\\ &=2\left(1+\frac{h}{64}+O(h^2)\right)-2\\\\ &=\frac h{32}+O(h^2) \tag 1 \end{align}$$

Dividing $(1)$ by $h$ and letting $h\to 0$ recovers the expected result.

Answer 2

Note that $2 = 16^{\frac{1}{4}}$. Hence this limit is the derivative of $f: x \mapsto x^{\frac{1}{4}}$ at the point $x=16.$ One has $$f^{\prime}(16) = \frac{1}{4} 16^{{\frac{-3}{4}}}= \frac{1}{32}.$$

Answer 3

Another solution is: \begin{equation} \lim_{h\to0}\frac{\sqrt[4]{16+h}-2}{h}=\lim_{h\to0}\frac{2\left(\displaystyle\frac{1}{2}\sqrt[4]{16+h}-1\right)}{h}=\lim_{h\to0}\frac{2\left(\sqrt[4]{\displaystyle1+\frac{h}{16}}-1\right)}{h}=\\ =2\lim_{h\to0}\frac{\left(\sqrt[4]{\displaystyle1+\frac{h}{16}}-1\right)}{\displaystyle16\frac{h}{16}}=\frac{2}{16}\cdot\frac{1}{4}=\frac{1}{32}. \end{equation}