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

[Loppyfoot]

Homework Statement

Find the limit:
lim (e^2x-1)/tanx
x$$\rightarrow$$0

 

[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$$\rightarrow$$0

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