I'm learning calculus right now and at current about limits. Please have a look at the image.Image 1

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.

  • $\begingroup$ Usually if you don't want use Hospital you should involve it's proof's details to your problem. $\endgroup$ – Hoseyn Heydari Nov 18 '15 at 20:49
  • 2
    $\begingroup$ First thing: Any proof starting out with "let A = lim", when you haven't shown the limit exists yet, is off to a bad start. Second: It's possible that $f(x)=0$ along a sequence of points tending to $a.$ In that case division by $f(x)$ makes no sense here. $\endgroup$ – zhw. Nov 18 '15 at 21:01
  • $\begingroup$ Have you covered derivatives at all? Do you know $\log'(x) = 1/x?$ $\endgroup$ – zhw. Nov 18 '15 at 21:49

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 $

  • $\begingroup$ Why “informal”? Since $\lim_{x\to\infty}{f(x)}=0$, we surely have $1+f(x)>0$ for $x>k$ (for some $k$). Or similarly, if the limit is at $a$. $\endgroup$ – egreg Nov 18 '15 at 20:53
  • $\begingroup$ @egreg Mainly informal because I'm assuming $f(x) \neq 0$ for any $x$, and for a couple other reasons zhw posted in comments above $\endgroup$ – Brenton Nov 18 '15 at 21:33
  • $\begingroup$ The full argument in the proof implicitly uses that $f(x)\ne0$ in a neighborhood of $a$. A rigorous proof of the statement can be obtained with Taylor developments, of course. And it's quite absurd to do so complicated limits before doing work on derivatives. $\endgroup$ – egreg Nov 18 '15 at 21:41

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 $x = a$.


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.


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 $.


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