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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?

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You've mistaken finding the derivative: for your function $f'(x)=e^{-x^2}$, so limit does exist. – CuriousGuest Jun 29 '14 at 15:18
I edited my post. – user159870 Jun 29 '14 at 15:19
Well, now the integral diverges on both ends! – Ted Shifrin Jun 29 '14 at 15:20
Allow me to tell a little story. When I was much younger, I read to my horror in a high-school math text that a differentiable function $f(x)$ has an asymptote for $x \to \infty$ iff $\lim_{x\to\infty} f'(x)$ exists. Now the examples here show that one implication is not true. The other is disproved by $f(x) = \sin(1/x)$. – Andreas Caranti Jun 29 '14 at 17:29

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

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will $\frac{1}{x} \sin x$ also work ? – Rene Schipperus Jun 29 '14 at 17:17
Nope, $(\sin(x)/x)' = -\sin(x)/x^{2} + \cos(x)/x$ goes to zero as $x \to \infty$. – Andreas Caranti Jun 29 '14 at 17:21
Nice, thanks... – Rene Schipperus Jun 29 '14 at 17:22
@ReneSchipperus, you're welcome! – Andreas Caranti Jun 29 '14 at 17:22

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

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So,is my attempt wrong? Why? – user159870 Jun 29 '14 at 15:25
As Ted Shifrin said in the comment to your question, integral of $e^{t^2}$ diverges. – CuriousGuest Jun 29 '14 at 15:26
To elementarily prove that $\lim_{x\to \infty} f(x)$ exists for the above function (without the residue theorem), we can break the integral up into an alternating sum and apply the alternating series test. – whosleon Jun 29 '14 at 15:30

Without using integrals, you can consider

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