Prove the tangent line exists

Question

How would one prove that the tangent line at $4$ exists on the function $f(x) = \frac{1}{x}+ \sqrt{x}$ using the formula: $$m=\lim_{x\to 0}\frac{\frac{1}{4+h}+\sqrt{4+h} - \frac{9}{4}}{h}$$

I can't wrap my head around this one :/.

Thanks in advance.

Answer 1

$$\lim_{h\rightarrow0}\frac{\frac{1}{4+h}+\sqrt{4+h}-\frac{9}{4}}{h}=\lim_{h\rightarrow}\frac{\frac{1}{4+h}-\frac{1}{4}}{h}+\lim_{h\rightarrow0}\frac{\sqrt{4+h}-2}{h}=$$ $$=\lim_{h\rightarrow}\frac{-h}{4h(4+h)}+\lim_{h\rightarrow0}\frac{h}{h(\sqrt{4+h}+2)}=-\frac{1}{16}+\frac{1}{4}=\frac{3}{16}$$

Answer 2

It will be easier to calculate if you know what is derivative and derivative theorems.

Definition of derivative:

For a real-valued function $f$ with domain the subset of $\mathbb R$, consider $x\in dom(f)$, if $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$exist and is finite (i.e.$\in\mathbb R$), then $f$ is differentiable at $x$ and the limit is defined as $f'(x)$.

(Some of) Derivative theorems:

If $f$,$g$ are differentiable at $x_0$, then $(f+g)'(x_0)=f'(x_0)+g'(x_0)$

If If $f$,$g$ are differentiable at $x_0$ and $g(x_0)\neq0$, then $(\frac{f}{g})'(x_0)=\frac{g(x_0)f'(x_0)-f(x_0)g'(x_0)}{g^2(x_0)}$

If $r\in\mathbb Q$. If $f(x)=x^r$ on its natural domain (all $x$ such that $f(x)$ make sense), then forall $x_0\in dom(f)$ and $x_0\neq0$, $f'(x_0)=rx^{r-1}$
(For its natural domain, for $r\neq0$ (obvious for $r\neq0)$, let $r=\frac{m}{n}$ in its reduced form, while $m\in\mathbb Z, n\in\mathbb N$, then consider $f=g\circ h$, while $g(x)=x^{1/n}$ and $h(x)=x^m$.)

If $f(x)=x, f'(x)=1$

To the question:

The question ask for $f'(4)$, while $f=\frac{1}{x}+\sqrt x$, can you continue from here?

Notes: For an indepth study of differential calculus, I will recommend Elementary Analysis: The theory of calculus (Edition 2) By Kenneth A. Ross. I believe it is available if you google search it.