Limits and derivative: is this proof accurate enough?

[Felafel]

Homework Statement

f is differentiable in $\mathbb{R^+}$ and
$\displaystyle \lim_{x \to \infty} (f(x)+f'(x))=0$
Prove that
$\displaystyle \lim_{x \to \infty}f(x)=0$

The Attempt at a Solution

I can split the limit in two:
$(\displaystyle \lim_{x \to \infty} f(x)+\displaystyle \lim_{x \to \infty} f'(x))=0$
I consider the second one and say that, by definition of derivative I have:
$\displaystyle \lim_{x \to \infty} \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$
As f is differentiable, then the second limit exists and is 0.
So, i have $\displaystyle \lim_{x \to \infty} 0 =0$
And then, by hypothesis:
$\displaystyle \lim_{x \to \infty} (f(x)+0)=0$
Are the passages logically correct?
thank you in advance!

 

[micromass]

Homework Statement

f is differentiable in $\mathbb{R^+}$ and
$\displaystyle \lim_{x \to \infty} (f(x)+f'(x))=0$
Prove that
$\displaystyle \lim_{x \to \infty}f(x)=0$

The Attempt at a Solution

I can split the limit in two:
$(\displaystyle \lim_{x \to \infty} f(x)+\displaystyle \lim_{x \to \infty} f'(x))=0$

No, you can't do that. You can only do that if you know that both limits exist. So:

$$\lim_{x\rightarrow +\infty} f(x)+g(x) = \lim_{x\rightarrow +\infty} f(x) + \lim_{x\rightarrow +\infty} g(x)$$

is not true in general, but only if you know that the limits actually exist.

As a counterexample:

$$0=\lim_{x\rightarrow +\infty} x-x \neq \lim_{x\rightarrow +\infty} x + \lim_{x\rightarrow +\infty} -x$$

 

[Felafel]

No, you can't do that. You can only do that if you know that both limits exist. So:

$$\lim_{x\rightarrow +\infty} f(x)+g(x) = \lim_{x\rightarrow +\infty} f(x) + \lim_{x\rightarrow +\infty} g(x)$$

is not true in general, but only if you know that the limits actually exist.

As a counterexample:

$$0=\lim_{x\rightarrow +\infty} x-x \neq \lim_{x\rightarrow +\infty} x + \lim_{x\rightarrow +\infty} -x$$

Thanks, you saved me from a major mistake!
Maybe I should prove this way, then?:

By definition of derivative:
$f(x)+f'(x)=f(x)+ \displaystyle \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$=

$\displaystyle \lim_{h \to 0} f(x)-\frac{f(x)}{h}+\frac{f(x+h)}{h}$ =
$\displaystyle \lim_{h \to 0} f(x)(1-\frac{1}{h})+\frac{f(x+h)}{h}$=

and, by hypothesis:

$\displaystyle \lim_{x \to \infty} \displaystyle \lim_{h \to 0} f(x)(1-\frac{1}{h})+\frac{f(x+h)}{h}$= 0

Say:

$\displaystyle \lim_{x \to \infty}f(x)=L$ so that the expression above becomes:

$\displaystyle \lim_{h \to 0}L(1-\frac{1}{\epsilon})+\frac{L}{\epsilon}$=0

$\displaystyle \lim_{h \to 0} L=0$ $\Rightarrow$ $\displaystyle \lim_{x\to \infty }f(x)=L$=0

 

[micromass]

Now you assume that

$$\lim_{h\rightarrow 0} \lim_{x\rightarrow +\infty} f(x,h) = \lim_{x\rightarrow +\infty}\lim_{h\rightarrow 0} f(x,h).$$

This is also not true in general.

 

[Felafel]

Now you assume that

$$\lim_{h\rightarrow 0} \lim_{x\rightarrow +\infty} f(x,h) = \lim_{x\rightarrow +\infty}\lim_{h\rightarrow 0} f(x,h).$$

This is also not true in general.

oh.. I've run out of possible good ideas then, any hint?

 

[micromass]

The intuition is this: if x is very large, then f(x) is very close to -f'(x). In particular, if f(x) is very large, then -f'(x) is very negative. And thus f(x) decreases very fast.

Try to work with an epsilon-delta definition. What does it mean that f(x) does not tend to 0??

 

[Felafel]

can I simply say then, that:
$\forall \epsilon >0$$\exists \delta>0$ such that if $0<|x-x_0|<\delta \Rightarrow |f(x)+f'(x)-0|<\epsilon$
And so, being the function >0 because it is defined on $\mathbb{R^+}:$0<|f(x)|< \epsilon - f'(x)|## $\Rightarrow$ $|f(x)|< \epsilon$ and for the squeeze rule $\lim f(x)=0$