How to prove $\lim_{x \to \infty} \frac{\log(1+f(x))}{f(x)} = 1$ without using L'Hopital's rule? I'm learning calculus right now and at current about limits. Please have a look at the image.
I can't understand the expression $\lim_{x \to a} \frac{\log(1+f(x))}{f(x)} = 1$ which is written under the bracket. Can anyone help me out how can I prove it.
EDIT: I want to know the proof without using L'Hopital's rule. 
 A: Here's an informal way to see intuitively why it is true for $f(x)>0$. We will use the identity $1 - \frac{1}{x} \leq \log x \leq x -1$ which is true for $x>0$ and are fairly well-known.
Replacing $x$ by $1+f(x)$:
$1 - \frac{1}{f(x)+1} \leq \log (1+ f(x)) \leq f(x)+1-1$
Dividing through by $f(x)$:
$\frac{1}{f(x)} - \frac{1}{f(x)(f(x)+1)} \leq \frac{\log (1+ f(x))}{f(x)} \leq 1$
But the left hand side simplifies down to:
$\frac{f(x)}{f(x)(1+f(x))} = \frac{1}{1+f(x)}$
Since $\lim_{x \rightarrow a} f(x) = 0$, you get:
$1 \leq \lim_{x \rightarrow a} \frac{\log (1+ f(x))}{f(x)} \leq 1 $
A: The limit $$\lim_{x \to a}\frac{\log(1 + f(x))}{f(x)} = 1$$ is true if the following condition holds:
$\lim_{x \to a}f(x) = 0$ and $f(x) \neq 0$ in some deleted neighborhood of $x = a$.
Why??
This is an easy consequence of the standard limit $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1\tag{1}$$ and the rule of substitution of limit below:
If $\lim_{x \to a}g(x) = b, \lim_{x \to b}f(x) = L$ and $g(x) \neq b$ in some deleted neighborhood of $x = a$ then $\lim_{x \to a}f(g(x)) = L$.
I suppose you want to know the proof of the standard limit $(1)$. This is not difficult but requires a proper definition of $\log x$. The simplest is to use the definition $$\log x = \int_{1}^{x}\frac{dt}{t}\tag{2}$$ which gives us $$\frac{d}{dx}\log x = \frac{1}{x}\tag{3}$$ using Fundamental Theorem of Calculus. Hence if $F(x) = \log x$ then $F'(1) = 1$ and therefore $$\lim_{h \to 0}\frac{F(1 + h) - F(1)}{h} = 1$$ or $$\lim_{h \to 0}\frac{\log(1 + h)}{h} = 1$$ Another definition of $\log x$ is as follows $$\log x = \lim_{n \to \infty}n(x^{1/n} - 1)\tag{4}$$ Using this definition to prove $(1)$ is bit tricky and given here. The main idea is to use the definition $(4)$ to establish the inequality $$1 - \frac{1}{x} \leq \log x \leq x - 1$$ for $x \geq 1$ and then proceed as in Brenton's answer.
A: If you know that
$\log(x)
=\int_1^x \frac{dt}{t}
$,
then
$\log(1+x)
=\int_1^{1+x} \frac{dt}{t}
=\int_0^{x} \frac{dt}{1+t}
$.
Since
$1 \le 1+t
\le 1+x$
for
$0 \le t \le x$,
$1 \ge \frac1{1+t}
\ge \frac1{1+x}$.
Integrating this
from $0$ to $x$,
$\log(1+x)
\le x
$
and
$\log(1+x)
\ge \frac{x}{1+x}
$.
If $0 \le x < 1$,
since
$1
\ge 1-x^2
=(1+x)(1-x)
$,
$\frac{1}{1+x}
\ge 1-x
$
so that
$\frac{x}{1+x}
\ge x(1-x)
=x-x^2
$.
Therefore
$\log(1+x)
\ge \frac{x}{1+x}
\ge x-x^2
$.
Finally,
if $0 \le x < 1$,
$x-x^2
\le \log(1+x)
\le x
$.
