Evaluate limits?

**demonmath** · November 2nd 2009, 05:24 PM

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.

**VonNemo19** · November 2nd 2009, 05:33 PM

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?

**demonmath** · November 2nd 2009, 06:06 PM

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

**VonNemo19** · November 2nd 2009, 06:09 PM

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?

**demonmath** · November 2nd 2009, 06:33 PM

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.

**VonNemo19** · November 2nd 2009, 06:40 PM

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 $\infty-\infty$ . The goal here is to rewrite this into a function that produces - by substitution - either $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , so that L'Hopital's rule will apply. Can you do this?

**demonmath** · November 2nd 2009, 07:03 PM

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

**VonNemo19** · November 2nd 2009, 07:05 PM

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...

**demonmath** · November 2nd 2009, 07:29 PM

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!

**VonNemo19** · November 2nd 2009, 07:34 PM

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

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

**demonmath** · November 2nd 2009, 07:42 PM

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

**mr fantastic** · November 2nd 2009, 07:47 PM

Originally Posted by VonNemo19

Sure.

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

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

A well known limit is $\lim_{u \to +\infty}\left(1+\frac{a}{u}\right)^{u} = e^a$ .

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

**demonmath** · November 2nd 2009, 07:58 PM

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

**mr fantastic** · November 2nd 2009, 11:36 PM

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 $a$ . Write down the value of your limit.

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