Let $f$ be a continuously differentiable function on $\mathbb{R}$. Suppose that $L=\lim_{x\to \infty}(f(x)+f'(x))$ exists.

If $ 0\lt L \lt\infty$ ,then which of the following statements is/are correct?

1.If $\lim_{x\to \infty}f'(x)$ exists then it is $0$

2.If $\lim_{x\to \infty}f(x)$ exists then it is $L$

3.If $\lim_{x\to \infty}f'(x)$ exists then $\lim_{x\to \infty} f(x) =0$

4.If $\lim_{x\to \infty}f(x)$ exists then $\lim_{x\to \infty}f'(x) =L$

Let $f(x)=1+ 1/x$ then $\lim_{x\to \infty}f(x)=1$ and $\lim_{x\to \infty}f'(x)=0$ so $L=1$

using this example 1,2 correct but this function is not differentiable at x=0 so how to proceed..

  • $\begingroup$ @TheDeadLegend To be pedantic, your two statements have no relationship. $\endgroup$
    – Kenny Lau
    May 11, 2017 at 13:18
  • 1
    $\begingroup$ This is a famous problem (if you search enough you will find it on this site). There is no need for continuous differentiability, just differentiability is sufficient. Also under these conditions the limit of $f'(x) $ will exist and be equal to $0$. So there is no need to write "if" limit of $f'$ exists. $\endgroup$
    – Paramanand Singh
    May 12, 2017 at 7:33

2 Answers 2


You made a very common, but also very bad, logical error:

You cannot prove a statement by providing an example, you can only disprove it.

For example:

My claim:

  • If a function is continuous, then it is bounded

My proof:

  • Take the example $f(x)=\sin x$. Clearly, $f$ is bounded, so my claim is correct.

Or even worse, my claim:

  • All white people live in Europe

My proof:

  • Take for example me, a white person. I live in Europe, so my claim is correct.

To actually prove the claim, you need to prove it for every function $f$.

To do that, here's a couple of hints:

  • If $\lim g(x)$ and $\lim h(x)$ both exist, then $\lim(g(x)-h(x))$ also exists.
  • From that, you can conclude that if $\lim g(x)+h(x)$ exists and $\lim g(x)$ exists, then $\lim h(x)$ exists (because $h(x)= (g(x)+h(x)) - g(x)$)
  • for $(1)$, think about what happens if the limit of $f'$ is not $0$.
  • for $(2)$, you can easily translate this into $(1)$.

If $\lim_{x\rightarrow\infty}f(x)$ exists and equals $l\in\mathbb{R}$ then $$l=\lim_{x\rightarrow\infty}f(x)=\lim_{x\rightarrow\infty}\frac{e^xf(x)}{e^x}=\lim_{x\rightarrow\infty}\frac{\frac{d}{dx}e^xf(x)}{\frac{d}{dx}e^x}=\lim_{x\rightarrow\infty}(f(x)+f'(x))$$ or $l=L$.

  • 1
    $\begingroup$ Nice trick, never heard of it before. $\endgroup$
    – 5xum
    May 11, 2017 at 13:19
  • $\begingroup$ i have neither. But the form $f(x)+f'(x)$ is so lucrative. :P $\endgroup$
    – QED
    May 11, 2017 at 13:52
  • $\begingroup$ There is no need to assume that limit of $f$ exists. L'Hospital's Rule proves that it exists and is equal to $L$ and $f'(x) \to 0$ as $x\to\infty$. $\endgroup$
    – Paramanand Singh
    May 12, 2017 at 7:35

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