Limit of (3^(sin(t))-1)/t as t approaches 0: L'hopital's problem

[SkyChurch]

Homework Statement

I need to find the lim as t approaches 0 of [(3^(sin(t)))-1]/t

Homework Equations

The Attempt at a Solution

I have no idea how to go about solving this. Whenever natural logs and "e" are involved I get confused. I know ln is involved here but I don't know how to go about using it. Thanks a lot!

 

[Dick]

You already said the magic word, l'Hopital. What the derivative of 3^(sin(t))? Remember 3^x=e^(ln(3)*x) and use the chain rule.

 

[SkyChurch]

so it would then be [(e^ln3*sint)-1]/t? I still don't know what to do next with e^(ln3*sint)? These ln and e problems really get me. How would I derive that term?

 

[NoMoreExams]

You should know what the derivative of $$e^{f(x)}$$ is if you are working with l'hospital's.

If you do not:

Spoiler

$$g(x) = e^{f(x)} \Rightarrow g'(x) = f'(x) e^{f(x)}$$

 

[SkyChurch]

Are you allowed to just give me the first derivative to that one term on these forums? Because that's the best way I learn things. I look at the derivative and can then just kind of figure out how you got there on my own and if I can't then I'll just ask questions, but I'm really not getting this one. Like I said natural logs and e give me a ton of trouble.

 

[NoMoreExams]

I won't speak for others but I don't think that's the best way to learn because on the exam you won't have that privilege. You can always buy REA Problem Solvers, Schaum's Outline, Calculus for Dummies, etc. Those books are GREAT if you learn by example as they have tons of them. Dick told you that

$$3^{sin(t)} = e^{ln(3^{sin(t)})} = e^{sin(t) \cdot ln(3)}$$

And I told you the rule to differentiate so where are you getting stuck?

 

[SkyChurch]

I'm getting stuck at deriving ln(3). I'm starting calc II right now so this stuff isn't very fresh in my mind.

 

[NoMoreExams]

ln(3) is a constant.

 

[SkyChurch]

Thanks a ton NME, I actually have calc for dummies but after a quick look through the workbook I couldn't find anything raised to the power of a trig function. I might be asking several more questions here in the upcoming days. I hate not understanding something, calc I didn't deal with ln and e enough for me to feel comfortable with those problems.

 

[NoMoreExams]

I would strongly suggest reviewing Algebra before you continue on. Most of Calculus 1 and 2 isn't hard, what trips a lot of people up is/are the hole(s) they have from previous classes. There's another way to do this problem which is equivalent and that's using implicit differentiation, i.e. take ln( ) of both sides.

 

[SkyChurch]

Yeah all of calc I was no problem. But I think in college algebra and pre calc they didn't spend enough time explaining the relationship of e and ln. Same with calc I. Every time I see a problem that involves or needs to involve ln or e I cringe.