For the following exercise use the formal definition to directly prove the f'(a) exists

Question

The definition in the book is: A real function f is said to be differentiable at a point $a \in \mathbb{R}$ if and only if f is defined on some open interval $I$ containing a and $$f'(a):=\lim_{x \to a} \frac{f(a+h)-f(a)}{h}$$

The function given is $$f(x) = \sqrt{x}, a\gt 0$$

$f'(a) = lim_{h \to 0} \frac{\sqrt{a+h}-\sqrt{a}}{h}$

I just don't know how to simplify from here. Super embarrassing.

Do you need to use the formal definition for the limit $L$ of $f(x)$ as $x$ approaches $c$: $$\lim_{x\to c}f(x)=L$$ if $$\forall\varepsilon>0,\ \exists\delta>0\ \text{such that}\ 0<|x-c|<\delta\implies |f(x)-L|<\varepsilon.$$ — Andrew Chin, Oct 24 '19 at 17:46

score 3 · Accepted Answer · answered Oct 24 '19 at 17:14

3

Just multiply the expression by $1=\frac{\sqrt{a+h}+\sqrt{a}}{\sqrt{a+h}+\sqrt{a}}$ to see that your differential quotient becomes $\frac{x+h-x}{h(\sqrt{a+h}+\sqrt{a})}=\frac{1}{\sqrt{a+h}+\sqrt{a}}\to\frac{1}{2\sqrt{a}}=f'(a)$.

answered Oct 24 '19 at 17:14

Duca_Conte

737

And why is $\lim_{h\to 0}\sqrt{a+h}=\sqrt{a}$? – Dr. Sonnhard Graubner Oct 24 '19 at 17:21
Use the method given in the answer of this other question : https://math.stackexchange.com/questions/2190573/prove-that-sqrtx-continuous-at-0 – Duca_Conte Oct 25 '19 at 08:19

score 0 · Answer 2 · answered Oct 24 '19 at 17:16

0

Write your term as $$\frac{\sqrt{a+h}-\sqrt{a}}{h}=\frac{a+h-a}{h(\sqrt{a+h}+\sqrt{a})}$$

answered Oct 24 '19 at 17:16

Dr. Sonnhard Graubner

95,283

For the following exercise use the formal definition to directly prove the f'(a) exists

2 Answers2