L'hospital's rule problem

**Afterme** · July 16th 2008, 01:23 PM

Hi guys I'm having trouble understanding this concept on problems like these

1. Lim CosX / 1 - Sinx
x->(pi/2)+
the answer would be Negative Infinity

2. Lim e^x - 1 / x^3
x->0
the answer would be Infinity

My question is why would the answers be infinity or negative infinity. I have no clue what is going on. Any detailed, step by step help would be greatly appreciated. Thanks

**Mathstud28** · July 16th 2008, 01:56 PM

Originally Posted by Afterme

Hi guys I'm having trouble understanding this concept on problems like these

1. Lim CosX / 1 - Sinx
x->(pi/2)+
the answer would be Negative Infinity

2. Lim e^x - 1 / x^3
x->0
the answer would be Infinity

My question is why would the answers be infinity or negative infinity. I have no clue what is going on. Any detailed, step by step help would be greatly appreciated. Thanks

For the first one Rewrite it as

$\lim_{x\to\frac{\pi}{2}}\frac{\sin\left(\frac{\pi} {2}-x\right)}{1-\cos\left(\frac{\pi}{2}-x\right)}$

Now let $t=\frac{\pi}{2}-x$

So as $x\to\frac{\pi}{2}\Rightarrow{t\to{0}}$

So we have

$\lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}$

Now $\sin(x)\sim{x}\text{ As }x\to{0}$

and

$1-\cos(x)\sim\frac{x^2}{2}\text{ As }x\to{0}$

So we can say that

$\lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}\sim\lim_{t\to{0}}\frac{t}{\frac{t^2}{2}}= \lim_{t\to{0}}\frac{2}{t}$

Now can you see why?

Or if you wanted to do L'hopital's

Since $\sin(0)=1-\cos(0)=0$

we have that

$\lim_{t\to{0}}\frac{\sin(t)}{1-\cos(t)}=\lim_{t\to{0}}\frac{\cos(t)}{-\sin(t)}$

Now we cannot apply L'hopital's again since it is not $\frac{0}{0}$ but $\frac{1}{0}$
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Now for the second one

$e^x-1\sim{x}\text{ As }x\to{0}$

So we have that

$\lim_{x\to{0}}\frac{e^x-1}{x^3}\sim\lim_{x\to{0}}\frac{x}{x^3}=\lim_{x\to{ 0}}\frac{1}{x^2}$

Now you should be able to see whats the deal there.

Or with L'hopitals

$\lim_{x\to{0}}\frac{e^x-1}{x^3}=\lim_{x\to{0}}\frac{e^x}{3x^2}$

Now we cannot apply Lhopitals again since this is $\frac{1}{0}$

**topsquark** · July 16th 2008, 03:26 PM

Originally Posted by Afterme

1. Lim CosX / 1 - Sinx
x->(pi/2)+
the answer would be Negative Infinity

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

This is of the form 0/0 so we can apply L'Hopital's rule.
$= \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}$

This is of the form -1/0, so the limit is at negative infinity.

-Dan

**Mathstud28** · July 16th 2008, 03:29 PM

Originally Posted by topsquark

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

This is of the form 0/0 so we can apply L'Hopital's rule.
$= \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}$

This is of the form -1/0, so the limit is at negative infinity.

-Dan

I am sorry, just out of curisoity, was there something wrong with my work?

**topsquark** · July 16th 2008, 03:31 PM

Originally Posted by Afterme

2. Lim e^x - 1 / x^3
x->0
the answer would be Infinity

This is of the form 0/0, so using L'Hopital's rule:
$\lim_{x \to 0} \frac{e^x - 1}{x^3} = \lim_{x \to 0} \frac{e^x}{3x^2}$

This is of the form 1/0, so it is positive infinity.

-Dan

**Afterme** · July 16th 2008, 06:18 PM

Originally Posted by topsquark

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

This is of the form 0/0 so we can apply L'Hopital's rule.
$= \lim_{x \to \frac{\pi }{2} ^2} \frac{-sin(x)}{cos(x)}$

This is of the form -1/0, so the limit is at negative infinity.

-Dan

Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.

**Mathstud28** · July 16th 2008, 07:01 PM

Originally Posted by Afterme

Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.

Think about it numerically

$\frac{-1}{.00000000\cdots{1}}=?$

**topsquark** · July 17th 2008, 06:05 AM

Originally Posted by Afterme

Hey Dan, thanks for the help but how does -1/0 become negative infinity? I thought -1/0 was an indeterminate answer.

Originally Posted by Mathstud28

Think about it numerically

$\frac{-1}{.00000000\cdots{1}}=?$

I'll back that up with the additional comment that "infinity" itself is an indeterminant answer.

-Dan

**Hahutzy** · July 17th 2008, 11:31 AM

It's because it's not really -1/0

Rather, it is -1/x where x is continuously approaching 0 but never reaching it.

Such number is hence continuously getting more negative, only bounded by the limit of negative infinity.

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