# Evaluate limits?

• Nov 2nd 2009, 06:24 PM
demonmath
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.
• Nov 2nd 2009, 06:33 PM
VonNemo19
Quote:

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?
• Nov 2nd 2009, 07:06 PM
demonmath
I'm going to say no...as i don't know for sure what that means.
• Nov 2nd 2009, 07:09 PM
VonNemo19
Quote:

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?
• Nov 2nd 2009, 07:33 PM
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.
• Nov 2nd 2009, 07:40 PM
VonNemo19
Quote:

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?
• Nov 2nd 2009, 08:03 PM
demonmath
i think i did...i just multiply to get a common denominator right?
• Nov 2nd 2009, 08:05 PM
VonNemo19
Quote:

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...
• Nov 2nd 2009, 08:29 PM
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!
• Nov 2nd 2009, 08:34 PM
VonNemo19
Quote:

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..
• Nov 2nd 2009, 08:42 PM
demonmath
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, 08:47 PM
mr fantastic
Quote:

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 $u = \frac{x}{5}$: $\lim_{u\to +\infty}\left(1+\frac{4/5}{u}\right)^{u}$.
• Nov 2nd 2009, 08:58 PM
demonmath
Not as obvious as I would have hoped, clearly. Im still stuck.
• Nov 3rd 2009, 12:36 AM
mr fantastic
Quote:

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.