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Can anybody check my proof? 
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#1
Nov812, 03:31 AM

P: 248

Let f be differentiable on some interval (c, infinity) and suppose that [tex] lim_{x \rightarrow \infty} [f(x) + f'(x)] = L [/tex], hwere L is finite. Prove that [tex]lim_{x \rightarrow \infty} f(x) = L[/tex] and [tex] lim_{x \rightarrow \infty} f'(x) = 0 [/tex]. Hint: [tex] f(x) = \frac{f(x)e^x}{e^x}[/tex]
My answer: For [tex]f(x) = \frac{f(x)e^x}{e^x}[/tex], Let h(x)=f(x)e^x and let g(x)=e^x. So we have [tex]f(x) = \frac{h(x)}{g(x)}[/tex] 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 [itex]ke^x[/itex] 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? 


#2
Nov812, 05:33 AM

P: 939

Or if you want an easier approach, that hint suggests to me that you might want to use L'Hopital. 


#3
Nov812, 07:11 AM

P: 206

The limit of a sum may be finite even if neither limit is finite.Take x and x.



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