Limits with L'Hopital

**totalnewbie** · December 17th 2006, 07:48 AM

Need help with these Hospital limits:

**CaptainBlack** · December 17th 2006, 08:11 AM

Find: $\lim_{x\to 0}\tan(x)\,\ln(x)$ .

Rewrite as:

$\lim_{x\to 0}\frac{\ln(x)}{\cot(x)}$ ,

now use L'hopital's rule:

$\lim_{x\to 0}\frac{\ln(x)}{\cot(x)}=\lim_{x \to 0}\frac{1/x}{(-\csc^2(x))}$ ,

...............= $\lim_{x \to 0}-\frac{\sin^2(x)}{x}$

...............= $0$

RonL

**CaptainBlack** · December 17th 2006, 08:15 AM

Find $\lim_{x \to 0}\frac{1}{x^2}-\frac{\sin(x)}{x}$

The first term in the limit clearly goes to $\infty$ , while the second
term is a well known limit equal to $1$ , so the limit as a whole is $\infty$ .

Opps didn't see the actual question:

Find $\lim_{x \to 0}\frac{1}{x^2}-\frac{\cos(x)}{\frac{\sin(x)}{x}}$

but the second term still has a finite limit and so the whle thing still goes to $\infty$

(or is a different interpretation of the second term in the limit meant?)

RonL

**Soroban** · December 17th 2006, 08:19 AM

Hello, totalnewbie!

Here's the first one . . .

$1)\;\lim_{x\to0}\left(\tan x\cdot\ln x\right)$

The function goes to: $(0)(-\infty)$

Rewrite as: . $\lim_{x\to0}\frac{\ln x}{\cot x}$ . . . which goes to $\frac{-\infty}{\infty}$

Apply L'Hopital: . $\lim_{x\to0}\left(\frac{\frac{1}{x}}{-\csc^2x}\right)\;=\;\lim_{x\to0}\left(-\frac{\sin^2x}{x}\right)$

. . $= \;\lim_{x\to0}\left[(-\sin x)\left(\frac{\sin x}{x}\right)\right] \;= \;-(0)(1) \;=\;\boxed{0}$

**totalnewbie** · December 17th 2006, 09:05 AM

I used Hospital rule for the second one too and got that final result is cosx*sinx. 0*1=0
Just want to see someone's confirmation.

**CaptainBlack** · December 17th 2006, 09:31 AM

Third go at the second question:

Find $\lim_{x \to 0} \frac{1}{x^2}-\frac{\cos(x)/\sin(x)}{x}$

which seems a plausible form for what might have been intended.

(the limit is 1/3 but I will need to do some work to produce what you would
think an acceptable explanation)

The limit is:

$\lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}$

Take the series expasion of $\cot(x)=\frac{1}{x}-\frac{x}{3}+O(x^3)$

So:

$\lim_{x \to 0} \frac{1}{x^2}-\frac{\cot(x)}{x}=\lim_{x \to 0}\frac{1}{x^2}-(1/x)\left[\frac{1}{x}-\frac{x}{3}+O(x^3)\right]$

.............. $=\lim_{x \to 0}\frac{1}{3}-O(x^2)=\frac{1}{3}$

RonL

**Soroban** · December 17th 2006, 09:40 AM

Hello, totalnewbie!

I am assuming that the second problem is: . $\lim_{x\to0}\frac{1}{x^2} - \frac{\frac{\cos x}{\sin x}}{x}$

Then we have: . $\frac{1}{x^2}-\frac{\cos x}{x\sin x} \;=\;\frac{\sin x - x\cos x}{x^2\sin x}$ . . . which goes to $\frac{0}{0}$

Apply L'Hopital: . $\frac{\cos x + x\sin x - \cos x}{x^2\cos x + 2x\sin x} \:=\:\frac{x\sin x}{x^2\cos x + 2x\sin x}$ $= \:\frac{\sin x}{x\cos x + 2\sin x}$

Divide top and bottom by $\sin x:\;\;\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}$

Then: . $\lim_{x\to0}\left(\frac{1}{\frac{x}{\sin x}\!\cdot\!\cos x + 2}\right) \;=\;\frac{1}{1\!\cdot1 + 2} \;=\;\boxed{\frac{1}{3}}$

**totalnewbie** · December 17th 2006, 11:28 AM

Thank you Soroban!

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