Show that it is possible that the limit $\displaystyle{\lim_{x \rightarrow +\infty} f'(x)} $ does not exist.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ a differentiable function with continuous derivative and the limit $\displaystyle{\lim_{x \rightarrow +\infty} f(x) }$ exists. Show with an example that it is possible that the limit $\displaystyle{\lim_{x \rightarrow +\infty} f'(x)} $ does not exist.

My attempt:

$$f(x)=\int_{-\infty}^{x} e^{t^2}dt$$

$$\lim_{x \to +\infty} f(x)=\lim_{x \to +\infty} \int_{-\infty}^{+\infty} e^{t^2}dt=\frac{\sqrt{\pi}}{2}$$

$$\lim_{x \to +\infty} f'(x) =\lim_{x \to +\infty} e^{x^2}= +\infty \notin \mathbb{R}$$

Is my attempt right?

Solution 1:

Another example, which used to be the standard one in my time, is $$ f(x) = \frac{1}{x} \sin( x^{2} ), $$ where $$ f'(x) = -\frac{1}{x^{2}} \sin(x^{2} ) + \frac{1}{x} 2 x \cos(x^{2}) = -\frac{1}{x^{2}} \sin(x^{2} ) + 2 \cos(x^{2}). $$

Solution 2:

Almost right: try $$f(x)=\int_0^x \sin(t^2)\,dt.$$

Solution 3:

Without using integrals, you can consider

• $\displaystyle \frac{\sin(x^2)}{x}$

• $\displaystyle \frac{\sin(x^3)}{x}$

• also this https://math.stackexchange.com/a/788818/66096