can a function be differentiable at a point if the difference quotient has a finite limit but the derivative has no limit?

Question

In the case of x^2*sin((5x+4)/x). the difference quotient has limit 0 for x->0, but the derivative itself is 2x*sin((5x+4)/x)+ 4*cos((5x+4)/x), which has no limit for x->0. So is the function differentiable at 0?

It is differentiable at $0$ but not continuously – marwalix Jan 27 '17 at 11:02 — marwalix, Jan 27 '17 at 11:02
Thats not true ! See my answer. – Fred Jan 27 '17 at 11:07 — Fred, Jan 27 '17 at 11:07

score 2 · Answer 1 · answered Jan 27 '17 at 11:07

2

I suppose that you have the following function:

$f(x)=x^2*\sin((5x+4)/x)$ for $x \ne 0$ and $f(0)=0$

Then

$| \frac{f(x)-f(0)}{x-0}|=|x|*|\sin((5x+4)/x)| \le |x|$,

therefore

$\frac{f(x)-f(0)}{x-0} \to 0$ for $x \to 0$.

This shows that $f$ is differentiable at $x=0$.

Furthermore the limit $\lim_{x \to 0}f'(x)$ does not exist. This means that $f'$ is not continuous at $x=0$.

answered Jan 27 '17 at 11:07

Fred

77,394

FWIW: We say that $f$ is differentiable but not "continuously differentiable". – ThomasR Jan 27 '17 at 14:07
This was exactly my comment – marwalix Jan 27 '17 at 16:16

can a function be differentiable at a point if the difference quotient has a finite limit but the derivative has no limit?

1 Answers1