# prove that $g(x) = \sqrt{x}, x\in [0, \infty]$ is differentiable

How to prove that $g(x) = \sqrt{x}, x\in [0, \infty]$ is differentiable?

Is my proof correct?

Let $x_0 \in [0, \infty]$.

\begin{align} \lim_{h \downarrow 0} \frac{g(x_0 + h) - g(x_0)}{h} &= \lim_{h \downarrow 0} \frac{\sqrt{x_0 + h} - \sqrt{x_0}}{h} \\ &= \lim_{h \downarrow 0} \frac{ \sqrt{x_0} - \sqrt{x_0} }{h} \\ &= 0 \\ &= \lim_{h \uparrow 0} \frac{ \sqrt {x_0 + h} - \sqrt{x_0}}{h} \\ &= \lim_{h \uparrow 0} \frac{g(x_0 + h) - g(x_0)}{h} \\ \end{align}

• can you assume continuity? – ℋolo Jan 2 '18 at 12:17
• correct me if I am wrong: if a function is differentiable on some domain then it implies that it is continous on that domain – jublikon Jan 2 '18 at 12:19
• You have just shown that $g'(x)=0\ \forall x \in [0,\infty]$. That's hideously wrong. More, your function has an undefined derivative at $x=0$ – Caffeine Jan 2 '18 at 12:20
• Your proof is not correct. How do you justify the second equality? – mathcounterexamples.net Jan 2 '18 at 12:21
• @jublikon yes, but you need to do more work, $\sqrt{x_0+h}\to \sqrt{x_0}$ if you know continuity. Another note is that it is goes to $\sqrt {x_0}$ so $\sqrt{x_0+h}-\sqrt{x_0}=L\to0$ and you can't say $L/h=0$ – ℋolo Jan 2 '18 at 12:26

Yes, but you need $x>0$ and to continue: $$\lim_{h\rightarrow0^+}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\rightarrow0^+}\frac{(\sqrt{x+h}+\sqrt{x})(\sqrt{x+h}-\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}=$$ $$=\lim_{h\rightarrow0^+}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=\lim_{h\rightarrow0^+}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2\sqrt{x}}.$$

• Can you explain your steps a bit more detailed please? That would be very great. I have some trouble to understand your equation – jublikon Jan 2 '18 at 12:25
• @jublikon I added something. See now. I used $(a-b)(a+b)=a^2-b^2$. – Michael Rozenberg Jan 2 '18 at 12:28
• @jublikon multiply and divide by $\sqrt{x+h}+\sqrt x$ – ℋolo Jan 2 '18 at 12:28

Having a glance to the result, we see you ve just proved that g(x) is a constant (since its derivative is 0 on a connected domain), and that's obviously not true (your mistake is that if the numerator goes to zero, that doesn't mean the wholr fraction goes to zero. In fact, if a function is derivable, the fraction has to present an undecision form: $\frac{0}{0}$ (which is not necessarily $0$). That's why a differentiable function is continuous: because $\lim_{h \to 0} f(x+h)-f(x)=0$

Doing in a correct way your calculation, we get that for $x>0$:

$$$g'(x)=\lim_{h \to \infty}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h \to \infty}\frac{h}{h(\sqrt{x+h}+\sqrt{h})}=\frac{1}{2x}$$$

Where the second equality is justified multiplying both numerator and denominator by $\sqrt{x+h}+\sqrt{x}$

Let $f$ be any continuous function on some interval $(a,b)$. Then, for $x_0\in(a,b)$, $$\lim_{h\to0}f(x_0+h)=f(x_0)$$ That's continuity, isn't it? Well some take this as the definition, others as a consequence, but it doesn't really matter.

Now, if your argument for the square root were sound, it would be also for our arbitrary $f$: $$\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h} \overset{\color{red}{!!!}}= \lim_{h\to0}\frac{f(x_0)-f(x_0)}{h}=0$$ (the three red exclamation marks point to the wrong step) so the derivative of any continuous function would be everywhere zero. This would make derivatives quite boring stuff.

You aren't allowed to replace only part of a function with its limit, when computing limits. Also the following one is incorrect: $$\lim_{x\to1}\frac{x^2-1}{x-1}\overset{\color{red}{!!!}}= \lim_{x\to1}\frac{1-1}{x-1}=0$$ According to this line of reasoning, every limit in the “indeterminate form” $0/0$ would be $0$, which it generally isn't: the right answer for the above limit is $2$: $$\lim_{x\to1}\frac{x^2-1}{x-1}=\lim_{x\to1}\frac{(x-1)(x+1)}{x-1} =\lim_{x\to1}(x+1)=2$$