# Can a function be differentiable if the limit does not exist?

I'm trying to find if the function is differentiable at $$x=1$$. Upon solving, the limit does not exist. But when I tried to solve for the limit of $$f '(1)$$, I both got $$3\over2$$, so $$f '(1)$$ exists. So, is the function differentiable at $$x=1$$? $$f(x) = \begin{cases} \ln(x^2+x) & x ≤ 1 \\ 3\sqrt x & x > 1 \\ \end{cases}$$

I. $$f(1) = \ln (1^2 +1) = \ln(2)$$ II. $$\lim_{x\to1^-} f(x) = \lim_{x\to1^-} \ln(x^2+x) = \ln(1^2+1) = \ln(2)$$ $$\lim_{x\to1^+} f(x) = \lim_{x\to1^+} 3 \sqrt x= 3\sqrt1 = 3$$ $$\lim_{x\to1^-} f(x) ≠ \lim_{x\to1^+} f(x)$$ Thus the $$\lim_{x\to1}f(x)$$ does not exist.

$$f'(x) = \begin{cases} \frac{2x+1}{x^2+x} & x < 1 \\ \frac{3}{2\sqrt x} & x > 1 \\ \end{cases}$$

a) $$f'_+(1) = \lim_{x\to1^+}f'(x )= \lim_{x\to1^+}\frac{3}{2\sqrt x} = \frac{3}{2\sqrt1} = \frac{3}{2}$$ b)$$f'_-(1) = \lim_{x\to1^-}f'(x)= \lim_{x\to1^+}\frac{2x+1}{x^2+ x} = \frac{2 + 1}{1+1} = \frac{3}{2}$$

• It is easier for people to help you if you show us what you have done.
– Jay
Mar 28, 2022 at 17:09
• It's in the pic. Mar 28, 2022 at 17:13
• math.meta.stackexchange.com/questions/9959/… Mar 28, 2022 at 17:23
• From our formatting and writing guidelines: "Don't force someone to click on an external link just to see or understand your question, it should be immediately visible after clicking on your title." Mar 28, 2022 at 17:30

You have not shown that $$f'(1)$$ exists. You have shown that $$\lim_{x\to1^+} f'(x) = \lim_{x\to1^-} f'(x),$$ but this does not imply that $$f'(1)$$ exists, or is equal to this limit.

To show that $$f'(1)$$ exists, you would have to show that the limit $$\lim_{h\to0} \frac{f(1+h) - f(1)}{h}$$ exists. But it is not too hard to see that this limit does not exist as $$h \to 0^+$$, as the numerator approaches $$3 - \ln 2 \neq 0$$ in this limit while the denominator will approach zero.

In general, a function must be continuous at any point at which it is differentiable (though the converse is of course not true.)

• Ooh, I get it now. Thanks a lot! Mar 28, 2022 at 17:22
• I think you've got a typo in the next-to-last paragraph: the first "denominator" should be "numerator", right? Mar 29, 2022 at 6:17
• @DavidZ: Fixed, thanks! Mar 29, 2022 at 11:51

A simple example: let $$f(x) = \begin{cases} -1 & x ≤ 0 \\ 1 & x > 0 \\ \end{cases}$$

The left and right limits of $$f'$$ at $$x=0$$ both exist and are equal to the same value (namely, zero), but that doesn't mean that $$f'(0)$$ exists.

Differentiability of real functions of one variable implies continuity of the function in that point, since you verify the function is not continuous neither is differentiable