# Homework Help: More Limits

1. Jun 2, 2012

### Saitama

1. The problem statement, all variables and given/known data
Find the values of a,b & c so that
$$\lim_{x→0} \frac{ae^x-bcos(x)+ce^{-x}}{x\cdot sin(x)}=2$$

2. Relevant equations

3. The attempt at a solution
The only thing which came to my mind is that when we put x=0, it should be a 0/0 form since the limit has a finite value. Therefore, a-b+c=0.
I can't think of any other equation to find the values of a,b and c.

Last edited: Jun 2, 2012
2. Jun 2, 2012

### Infinitum

Hi again!!

You have to use the L'Hospital rule for this. Applying the rule once will give you the second relation between a,b,c. L'Hospital'ing it another time will give you yet another relation. 3 equations, 3 variables....

3. Jun 2, 2012

### Saitama

Lol, i forgot to mention, i am not allowed to use L'Hospital rule here.

4. Jun 2, 2012

### Infinitum

Bah, this is the seventh question I'm trying today without L'Hospital rule. Poor guy.

5. Jun 2, 2012

### Saitama

I too feel so poor when i am not allowed to use L'Hospital rule. This is the seventh question for you, i have been spending my whole day to solve the Limits question without L'Hospital rule. At last, when i get exhausted, i come to PF so as to get some help!

6. Jun 2, 2012

### Infinitum

Oh, ouch! Take a break?

I was pitying the person who made the rule, actually. No one bothers to use it

Anyway, back on topic.....(trying...)

7. Jun 2, 2012

### ehild

Can you use Taylor expansion?

ehild

8. Jun 2, 2012

### Saitama

We aren't yet taught this.
I know a very little about Taylor expansion but never used it.

9. Jun 2, 2012

### Bohrok

Are there any limits you know/can use with ex and e-x?

10. Jun 2, 2012

### Saitama

The one in my mind is $\lim_{x\to 0} \frac{e^x-1}{x}=1$.

11. Jun 2, 2012

### Curious3141

Well, you *have* to be able to use *something*. Otherwise, the problem can't be solved.

If you've been taught Taylor expansion, you should use it. Look up the series expansions for $e^x$ (from which you can immediately deduce that for $e^{-x}$, $\sin x$ and $\cos x$.

You'll need the actual terms up to and including $x^2$ since the denominator is $x\sin x$. The remaining terms can be grouped together as $O(x^3)$ for ease.

Separate out the terms and figure out what each rational expression needs to be at the limit.

12. Jun 2, 2012

### Saitama

No, we aren't taught Taylor expansion yet but i do know the series you are asking me.
Even if i expand and write $e^{-x}$, $e^{-x}$, $\sin x$ and $\cos x$, i can't take out the factor x^2 from the numerator.

13. Jun 2, 2012

### SammyS

Staff Emeritus
From that you can find $\displaystyle \lim_{x\to 0} \frac{e^{-x}-1}{x}$

$\displaystyle \lim_{x→0} \frac{a(e^x-1)+a-(a+c)\cos(x)+c(e^{-x}-1)+c}{x\sin(x)}$

Then, the identity, $\displaystyle 1-\cos(x)=2\sin^2\left(\frac{x}{2}\right)$ may get you nearly home.

14. Jun 2, 2012

### Saitama

$$\lim_{x\to 0} \frac{a(e^x-1)+c(e^{-x}-1)+(a+c)(1-cos(x))}{x\sin(x)}$$