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)$$ f'_-(1) = \lim_{x\to1^-}f'(x)= \lim_{x\to1^+}\frac{2x+1}{x^2+ x} = \frac{2 + 1}{1+1} = \frac{3}{2}$$
Best Answer
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.)