# Thread: evaluating a limit with ln

1. ## evaluating a limit with ln

Hi

So the limit is

lim x ( ln (x+5) - ln x)
x-> infinity

so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

Where do I go from here to evaluate the limit?

Thanks

Calculus beginner

2. Originally Posted by calcbeg
Hi

So the limit is

lim x ( ln (x+5) - ln x)
x-> infinity

so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

Where do I go from here to evaluate the limit?

Thanks

Calculus beginner
$\displaystyle \lim_{x \to \infty}\left(x (\ln(x+5)-\ln(x)\right) = \lim_{x \to \infty} \ln\left( \left(\dfrac{x+5}{x}\right)^x\right)$

Since $\displaystyle \lim_{x \to \infty} \left( \left(1+\dfrac{5}{x}\right)^x\right) = e^5$

your limit exists and has the value 5.

3. Originally Posted by calcbeg
Hi

So the limit is

lim x ( ln (x+5) - ln x)
x-> infinity

so everything goes to infinity as x goes to infinity but I don't think I can use l'Hospital's rule b/c it is not in quotient form.

Where do I go from here to evaluate the limit?

Thanks

Calculus beginner
$\displaystyle \mathop {\lim }\limits_{x \to \infty } x\left( {\ln \left( {x + 5} \right) - \ln x} \right) = \mathop {\lim }\limits_{x \to \infty } x\ln \left( {1 + \frac{5}{x}} \right) = \mathop {\lim }\limits_{x \to \infty } \ln {\left( {1 + \frac{5}{x}} \right)^x} =$

$\displaystyle = \left\{ \begin{gathered} x = 5t, \hfill \\ x \to \infty , \hfill \\ t \to \infty \hfill \\ \end{gathered} \right\} = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1} {t}} \right)^{5t}} = \ln {e^5} = 5.$

4. Hello, calcbeg!

Find the limit: .$\displaystyle \lim_{x\to\infty}x \bigg[\ln(x+5) - \ln x\bigg]$

We have: .$\displaystyle \lim_{x\to\infty}x\bigg[\ln(x+5) - \ln(x)\bigg] \;=\;\lim_{x\to\infty}\bigg[\frac{\ln(x+5) - \ln(x)}{\dfrac{1}{x}}\bigg] \quad\rightarrow\quad \frac{0}{0}$

Apply L'Hopital: .$\displaystyle \lim_{x\to\infty}\left[\frac{\frac{1}{x+5} - \frac{1}{x}}{-\frac{1}{x^2}}\right] \;=\;\lim_{x\to\infty}\frac{5x}{x+5}$

Divide top and bottom by $\displaystyle x\!:\;\;\lim_{x\to\infty}\frac{\frac{5x}{x}}{\frac {x}{x} + \frac{5}{x}} \;=\;\lim_{x\to\infty}\frac{5}{1+\frac{5}{x}} \;=\;\frac{5}{1+0} \;=\;5$