Limit of (e^2x-1)/tanx as x approaches 0

[Loppyfoot]

Homework Statement

Find the limit:
lim (e^2x-1)/tanx
x\rightarrow0

 

[Mark44]

This is the indeterminate form [0/0]. Do you know L'Hopital's Rule?

 

[Loppyfoot]

No, we have not learned this rule yet. So, since it is indeterminate form, the limit would be 0, or does not exist?

 

[Office_Shredder]

It means you need to be careful. Knowing it's of the form 0/0 only tells you that it could be any value, and you need another method of evaluating it than the standard split the limit up trick.

Basically what you have is

\lim \frac{e^{2x}-1}{sin(x)}*cos(x)

the cos(x) part stays near 1, so is fine. You need a trick to manipulate \frac{e^{2x}-1}{sin(x)} to make it more amenable to calculation, either through Taylor series around 0 or some other method

 

[Mark44]

If L'Hopital's Rule hasn't been presented yet, it's a safe bet that you don't know about Taylor's series or Maclaurin series yet. Other than these techniques, I don't have any advice. Is this a problem in the book, or is it one your instructor gave for extra credit? The best advice I can offer is to contact him/her for a hint.

 

[VietDao29]

Homework Statement

Find the limit:
lim (e^2x-1)/tanx
x\rightarrow0

Well, in limit problems, when dealing with ln, and exponential function, there are 2 well-known limits you should memorize (I know it should be somewhere in your book, and proving it is not very hard, you can give it a try if you want):

1. \lim_{x \rightarrow 0} \frac{e ^ x - 1}{x} = 1 (Limit 1)
   
2. \lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1 (Limit 2)

When working with sin(x), there's one well-known (the most well-known, I may say) limit:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1. (Limit 3)

Ok, so now, back to your problem. First, we can try to expand every terms out:

\lim_{x \rightarrow 0} \frac{e ^ {2x} - 1}{\tan(x)} &= \lim_{x \rightarrow 0} \left( {\color{red}\frac{e ^ {2x} - 1}{\sin(x)}} \times {\color{blue}\cos(x)} \right)

Pay attention to the red term, it's (0/0), hence one of the Indeterminate Forms, right? The blue term will tend to cos(0) = 1, which does not cause much troubles. So, here the problematic term is the red one.

Now, think of a way to make use of (Limit 1), and (Limit 3) to eliminate the Indeterminate Form 0/0. It should be easy if you know the trick. Well, let's see how far can you go. :)

 

[JG89]

Note that e^{2x} - 1 = \int_0^{2x} e^u du = 2xe^c for some c in between 0 and 2x, by the MVT of integral calculus. Thus \frac{e^{2x} - 1}{sinx} cosx = \frac{2xe^c}{sinx} cosx = \frac{x}{sinx} 2e^c cosx.

Remember that \frac{x}{sinx} goes to 1 for x approaching 0, so now let x go to 0 in the entire new expression. Also remember that c must also go to 0 as x goes to 0.

 

[Mark44]

Note that e^{2x} - 1 = \int_0^{2x} e^u du = 2xe^c for some c in between 0 and 2x, by the MVT of integral calculus. Thus \frac{e^{2x} - 1}{sinx} cosx = \frac{2xe^c}{sinx} cosx = \frac{x}{sinx} 2e^c cosx.

Remember that \frac{x}{sinx} goes to 1 for x approaching 0, so now let x go to 0 in the entire new expression. Also remember that c must also go to 0 as x goes to 0.

This advice might not be helpful to the OP if he is studying differential calculus and hasn't gotten to integral calculus yet.

 

[TheFurryGoat]

[*]\lim_{x \rightarrow 0} \frac{e ^ x - 1}{x} = 1 (Limit 1)

When working with sin(x), there's one well-known (the most well-known, I may say) limit:

\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1. (Limit 3)

e^2x - 1 = (e^x)^2 - 1 = (e^x - 1)(e^x + 1)

=> (e^2x - 1)/tanx = (e^x-1)/sinx * (e^x+1)cosx.
What can you do now to be able to use the known limits? For example multiply by 1=a/a=b/b=c/c=...

 

[lurflurf]

for x!=0
[exp(2x)-1]/tan(x)=[(exp(x)-1)/x][exp(x)+1][cos(x)]/[sin(x)/x]
you should know
exp'(0)=lim [(exp(x)-1)/x]=1
sin'(0)=lim [sin(x)/x]=1
exp(0)=1
cos(0)=1