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.

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- Nov 2nd 2009, 05:24 PMdemonmathEvaluate 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. - Nov 2nd 2009, 05:33 PMVonNemo19
- Nov 2nd 2009, 06:06 PMdemonmath
I'm going to say no...as i don't know for sure what that means.

- Nov 2nd 2009, 06:09 PMVonNemo19
- Nov 2nd 2009, 06:33 PMdemonmath
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.

- Nov 2nd 2009, 06:40 PMVonNemo19
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? - Nov 2nd 2009, 07:03 PMdemonmath
i think i did...i just multiply to get a common denominator right?

- Nov 2nd 2009, 07:05 PMVonNemo19
- Nov 2nd 2009, 07:29 PMdemonmath
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!

- Nov 2nd 2009, 07:34 PMVonNemo19
- Nov 2nd 2009, 07:42 PMdemonmath
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 - Nov 2nd 2009, 07:47 PMmr fantastic
- Nov 2nd 2009, 07:58 PMdemonmath
Not as obvious as I would have hoped, clearly. Im still stuck.

- Nov 2nd 2009, 11:36 PMmr fantastic