limit as x -> infty

**amthomasjr** · October 27th 2012, 07:38 PM

$\lim_{x \rightarrow \infty}(x-lnx)$

I made it into indeterminate form so that I can use l'Hospital's rule:
$\lim_{x \rightarrow \infty}(lne^x-lnx) = \lim_{x \rightarrow \infty}ln\left(\frac{e^x}{x}\right)$

I take the derivative and get:

$\frac{x-1}{x}$

Taking the derivative again I get that the function approaches 1, but based off of the actual graph the function approaches infinity. What am I doing wrong?

**MarkFL** · October 27th 2012, 08:02 PM

$\lim_{x\to\infty}\ln\left(\frac{e^x}{x} \right)=\ln\left(\lim_{x\to\infty}\frac{e^x}{x} \right)$

Now apply L'Hôpital's rule to the limit inside the log function.

**Soroban** · October 27th 2012, 08:05 PM

Hello, amthomasjr!

$\lim_{x \rightarrow \infty}(x-\ln x)$

I made it into indeterminate form so that I can use l'Hospital's rule:

$\lim_{x\to\infty}(\ln e^x-\ln x) \:=\: \lim_{x\to\infty}\ln\left(\frac{e^x}{x}\right)$

I take the derivative and get: . $\frac{x-1}{x}$ . How?

Exactly what did you differentiate?

**amthomasjr** · October 27th 2012, 08:16 PM

Originally Posted by Soroban

Hello, amthomasjr!

Exactly what did you differentiate?

$ln\left(\frac{e^x}{x}\right)$

**MarkFL** · October 27th 2012, 08:22 PM

In order to apply L'Hôpital's rule you need the limit to take the form:

$\lim_{x\to c}\frac{f(x)}{g(x)}$

**amthomasjr** · October 27th 2012, 08:22 PM

Originally Posted by MarkFL2

$\lim_{x\to\infty}\ln\left(\frac{e^x}{x} \right)=\ln\left(\lim_{x\to\infty}\frac{e^x}{x} \right)$

Now apply L'Hôpital's rule to the limit inside the log function.

That makes sense, so

$ln\left(e^x\right) = x$ and

$\lim_{x\to\infty}x = \infty$

**amthomasjr** · October 27th 2012, 08:23 PM

Originally Posted by MarkFL2

In order to apply L'Hôpital's rule you need the limit to take the form:

$\lim_{x\to c}\frac{f(x)}{g(x)}$

Which is why you only apply it to the inside, correct?

**MarkFL** · October 27th 2012, 08:31 PM

I would write:

$\ln\left(\lim_{x\to\infty}\frac{e^x}{x} \right)=\ln\left(\lim_{x\to\infty}e^x \right)=\ln(\infty)=\infty$

**MarkFL** · October 27th 2012, 08:34 PM

Originally Posted by amthomasjr

Which is why you only apply it to the inside, correct?

Right, on the inside of the log function, you now have the correct form to use the rule bought by L'Hôpital.

**amthomasjr** · October 27th 2012, 08:41 PM

Originally Posted by MarkFL2

Right, on the inside of the log function, you now have the correct form to use the rule bought by L'Hôpital.

I'm assuming the same thing applies to trig functions if there were a similar case

**MarkFL** · October 27th 2012, 08:56 PM

Yes, I believe so as long as the trig function is continuous.

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