# evaluating a limit with ln

• October 31st 2009, 11:18 AM
calcbeg
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
• October 31st 2009, 11:44 AM
earboth
Quote:

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

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

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

your limit exists and has the value 5.
• October 31st 2009, 11:47 AM
DeMath
Quote:

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

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

$= \left\{ \begin{gathered}
x = 5t, \hfill \\
x \to \infty , \hfill \\
t \to \infty \hfill \\
\end{gathered} \right\} = \mathop {\lim }\limits_{t \to \infty } \ln {\left( {1 + \frac{1}
{t}} \right)^{5t}} = \ln {e^5} = 5.$
• October 31st 2009, 12:11 PM
Soroban
Hello, calcbeg!

Quote:

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

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

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

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