Is there an example of a function $f:\mathbb{R}→ \mathbb{R}$, that is differentiable at $0$ but does not have a limit at $0$.
My initial thoughts was that this does not exist as the definition of a differentiable function includes limits. That is $\displaystyle\lim_{x →0} \frac{f(x)-f(0)}{x} $ so if this exists then surely does $\displaystyle\lim_{x→0} f(x)$. But I know do not think this argument follows.
Is there a way there to prove this or an example that does fit these constraints.
Your argument almost works. By adding some details, you end up proving that if $f$ is differentiable at $0$ then $f$ is continuous at $0$, and therefore $\lim_{x \to 0} f(x) = f(0)$.
Here's some additional detail.
Knowing that $f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x}$ exists, and knowing that $\lim_{x \to 0} x = 0$, by using the product theorem for limits it follows that $$\lim_{x \to 0} (f(x)-f(0)) = \lim_{x \to 0} \frac{f(x)-f(0)}{x} \cdot x $$ exists, and furthermore $$\lim_{x \to 0} (f(x)-f(0)) = \lim_{x \to 0} \frac{f(x)-f(0)}{x} \cdot x = \lim_{x \to 0} \frac{f(x)-f(0)}{x} \cdot \lim_{x \to 0} x = f'(0) \cdot 0 = 0 $$ Next, knowing that the limit of the constant $f(0)$ exists, and knowing the sum theorem for limits, it follows that $\lim_{x \to 0} f(x) = \lim_{x \to 0} ((f(x)-f(0)) + f(0))$ exists, and that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} ((f(x)-f(0)) + f(0)) = \lim_{x \to 0} (f(x)-f(0)) + \lim_{x \to 0} f(0) = 0 + f(0) = f(0) $$
