## Can anybody check my proof?

Let f be differentiable on some interval (c, infinity) and suppose that $$lim_{x \rightarrow \infty} [f(x) + f'(x)] = L$$, hwere L is finite. Prove that $$lim_{x \rightarrow \infty} f(x) = L$$ and $$lim_{x \rightarrow \infty} f'(x) = 0$$. Hint: $$f(x) = \frac{f(x)e^x}{e^x}$$

For $$f(x) = \frac{f(x)e^x}{e^x}$$, Let h(x)=f(x)e^x and let g(x)=e^x. So we have $$f(x) = \frac{h(x)}{g(x)}$$ Since we know that the sum of the limits of f(x) and f'(x) is finite, we know that each limit must also be finite. Therefore, h(x) eventually be less than g(x). So for large x, h(x) =< g(x) in order for h(x)/g(x) to converge. Since h(x)=f(x)e^x and g(x)=e^x, f(x) =< 1.

This can happen in two ways. Either h(x) is $ke^x$ where k is a constant such that |k|=< 1...

If this is the case, then the derivative of f(x) converges to zero, while f(x) itself converges to some number L.

...or f(x) (in f(x)e^x) is a decreasing function that converges to a number that is less than or equal to 1.

In this case, both f(x) and f'(x) converge to zero.

Do you think my answer is correct?
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 Quote by Artusartos Since we know that the sum of the limits of f(x) and f'(x) is finite, we know that each limit must also be finite.
It seems to me that this is exactly what you needed to show in the first place. Can you show this is true?

Or if you want an easier approach, that hint suggests to me that you might want to use L'Hopital.
 The limit of a sum may be finite even if neither limit is finite.Take x and -x.