1. ## L'hospital's rule problem

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

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

2. 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

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

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

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

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

So we have

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

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

and

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

So we can say that

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

Now can you see why?

Or if you wanted to do L'hopital's

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

we have that

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

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

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

So we have that

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

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

Or with L'hopitals

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

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

3. Originally Posted by Afterme
1. Lim CosX / 1 - Sinx
x->(pi/2)+
the answer would be Negative Infinity
$\displaystyle \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}$

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

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

-Dan

4. Originally Posted by topsquark
$\displaystyle \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}$

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

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?

5. Originally Posted by Afterme
2. Lim e^x - 1 / x^3
x->0
This is of the form 0/0, so using L'Hopital's rule:
$\displaystyle \lim_{x \to 0} \frac{e^x - 1}{x^3} = \lim_{x \to 0} \frac{e^x}{3x^2}$

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

-Dan

6. Originally Posted by topsquark
$\displaystyle \lim_{x \to \frac{\pi }{2} ^2} \frac{cos(x)}{1 - sin(x)}$

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

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.

7. 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.

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

8. 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
$\displaystyle \frac{-1}{.00000000\cdots{1}}=?$