# How to transform and simplify the limit : $\;\lim_{x\to 0} \frac{\ln(1+5x)}{x}$?

The result of the limit $$\;\lim_{x\to 0} \dfrac{\ln(1+5x)}{x}$$ should be $5$. How do I get it?

We have $$\text{for}\ x>0\quad\frac{5x}{1+5x}\le\ln(1+5x)=5\int_0^x\frac{dt}{1+5t}\le5x$$ so use the squeeze theorem to conclude.

Remark The limit on the right suffices to conclude since the function is continuous.

When evaluating a limit you want to let $x$ approach zero. For $f(x) = \frac{ln(1+5x)}{x}$ you get

$\lim_{x\rightarrow 0}f(x) = \frac{0}{0}$

at first glance. This should tip you off that you need to use L'Hopital's rule. L'Hopital's rule says that if your limit takes the form $\frac{0}{0}$ (or some other forms) you evaluate the limit by

$\lim_{x\rightarrow 0}f(x) = \lim_{x \rightarrow 0} \frac{\frac{d}{dx}ln(1+5x)}{\frac{d}{dx}x}$.

I will leave it to you to take the derivatives and evaluate them at $x=0$.

• Any other approach could be used than L'Hopital's rule? Nov 14, 2013 at 20:20
• Take a look at Sami's answer, it is valid if you are comfortable with generating inequalities. Nov 14, 2013 at 20:21
• The logic given here is circular. The L'Hospital technique is based on derivative of $\log x$ which can not be done unless we calculate the limit in question. You can not prove $A$ using $B$ if $B$ itself needs $A$ beforehand. This is one of the reason's I try to avoid use of L'Hospital. In most of the simple limit problems its use makes the argument circular. Nov 16, 2013 at 8:24

Let's keep in mind that $$\ln(1 + t) \sim t$$

Then

$$\frac{\ln(1+5x)}{x} = \frac{5x }{x} = 5$$

• Short and simple ! :-) Nov 14, 2013 at 20:52
• @Ant You need not multiply by $5$ just write $\ln(1+5x)\sim 5x$. Isn't it?
– user63181
Nov 14, 2013 at 21:06
• you're right, actually.. thanks ;-)
– Ant
Nov 15, 2013 at 6:46
• UpVote $0$ k. Fine. Nov 15, 2013 at 10:20

HINT: Use l'Hopitals rule to evaulate the limit.

$$\lim_{x\to0}\frac{\ln(1+5x)}x~=~\lim_{y\to\infty}\left[y\cdot\ln\left(1+\frac5y\right)\right]~=~\lim_{y\to\infty}\ln\left(1+\frac5y\right)^y~=~\lim_{y\to\infty}\ln\left(1+\frac1{y/5}\right)^{\dfrac y5~\cdot~\large5}~=$$
$$=~\lim_{t\to\infty}\ln\left(1+\frac1t\right)^{t~\cdot~5}~=~\ln\lim_{t\to\infty}\left(1+\frac1t\right)^{t~\cdot~5}~=~\ln\Big(e^5\Big)~=~5.$$
QED.