1. ## Evaluate limits?

Evaluate the following limits:

1) lim as x-->0+, x^(1/3)*ln(x)

2) lim as x-->1+, (1/ln(x) - 1/(x-1))

3) lim as x-->infinity, (1+(4/x))^(x/5)

I'm unsure how to even begin these ones...any help would be greatly appreciated.

2. Originally Posted by demonmath
Evaluate the following limits:

1) lim as x-->0+, x^(1/3)*ln(x)

2) lim as x-->1+, (1/ln(x) - 1/(x-1))

3) lim as x-->infinity, (1+(4/x))^(x/5)

I'm unsure how to even begin these ones...any help would be greatly appreciated.
Do you know your indeterminate forms?

3. I'm going to say no...as i don't know for sure what that means.

4. Originally Posted by demonmath
I'm going to say no...as i don't know for sure what that means.
It would help me if I knew if you were in Calc 1 or Calc 2. Are you supposed to be applying L'Hopital's Rule?

5. I'm in Calc 251, I'm not sure what the equivalent of that is. No instructions were given on the practice assignment question I posted except what I've included. There were four questions and I answered the first one using L'Hopitals, so I'm assuming there would be no problem with this.

6. Originally Posted by demonmath
I'm in Calc 251, I'm not sure what the equivalent of that is. No instructions were given on the practice assignment question I posted except what I've included. There were four questions and I answered the first one using L'Hopitals, so I'm assuming there would be no problem with this.
Ok. So, you instructer shouldv'e mentioned something about indeterminate forms, but anyway.

Not that by substitution, number 2 yields the indeterminate form $\displaystyle \infty-\infty$. The goal here is to rewrite this into a function that produces - by substitution - either $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, so that L'Hopital's rule will apply. Can you do this?

7. i think i did...i just multiply to get a common denominator right?

8. Originally Posted by demonmath
i think i did...i just multiply to get a common denominator right?
Indeed! then apply L'Hopital! See the magic happen...

9. i can get the first one now from your help as well but im struggling with the last one still. If you could lend a hand again that would be great!

10. Originally Posted by demonmath
i can get the first one now from your help as well but im struggling with the last one still. If you could lend a hand again that would be great!
Sure.

Try letting $\displaystyle y=\lim_{x\to\infty}\left(1+\frac{4}{x}\right)^{x/5}$

Now take the ln of both sides to get that nasty exponent out if there..

11. okay, so the exponent comes out front then. So should i get on the right side:

[ln(1+4/x)]/(5x^-1)

Or is there a better way to do it

12. Originally Posted by VonNemo19
Sure.

Try letting $\displaystyle y=\lim_{x\to\infty}\left(1+\frac{4}{x}\right)^{x/5}$

Now take the ln of both sides to get that nasty exponent out if there..
A well known limit is $\displaystyle \lim_{u \to +\infty}\left(1+\frac{a}{u}\right)^{u} = e^a$.

So the asked for limit is obvious if you make the substitution $\displaystyle u = \frac{x}{5}$: $\displaystyle \lim_{u\to +\infty}\left(1+\frac{4/5}{u}\right)^{u}$.

13. Not as obvious as I would have hoped, clearly. Im still stuck.

14. Originally Posted by demonmath
Not as obvious as I would have hoped, clearly. Im still stuck.
Look at the well known limit I pointed out to you. Look at what your limit becomes after the substitution is made. Compare the two. Identify the value of $\displaystyle a$. Write down the value of your limit.