How can you determine the following limit

[OmniNewton]

Homework Statement

How can you determine the following limit using rationalization?

where x = h

Homework Equations

The Attempt at a Solution

I attempted to multiply by the conjugate and cannot get the problem to work out. When doing this I noticed no terms canceled out and now I am stuck.

We have not use l'hopital's rule in our class and I am wondering how I can solve this problem without derivatives.
I simplified the limit to the following state:

Additionally, I tried taking the conjugate of the numerator as well. The problem is every substitution leads to 0/0

 

[axmls]

Do you mean the following?
$$\lim _{h \to 0} \frac{\sqrt{h^2 + 15} - \sqrt{h + 15}}{\sqrt{h + 3} - 2}$$

 

[OmniNewton]

Do you mean the following?
$$\lim _{h \to 0} \frac{\sqrt{h^2 + 15} - \sqrt{h + 15}}{\sqrt{h + 3} - 2}$$

yes

 

[SteamKing]

Homework Statement

How can you determine the following limit using rationalization? lim x--> 0 (h^2 + 15)^(1/2) - (h + 15)^(1/2) divided by (h+3)^1/2 - 2.

Homework Equations

The Attempt at a Solution

I attempted to multiply by the conjugate and cannot get the problem to work out. When doing this I noticed no terms canceled out and now I am stuck.

We have not use l'hopital's rule in our class and I am wondering how I can solve this problem without derivatives.

Is the limit supposed to be evaluated as x → 0 or as h → 0?

In the expression above, there is no x variable.

 

[OmniNewton]

Is the limit supposed to be evaluated as x → 0 or as h → 0?

In the expression above, there is no x variable.

Sorry that is a extremely simple mistake you are correct as h--->0

 

[axmls]

Have you tried simply plugging $h = 0$ in?

 

[OmniNewton]

Have you tried simply plugging $h = 0$ in?

My bad the problem is actually h----> 1 I have been looking at the problem for 2 hours :p

 

[axmls]

You don't necessarily need terms to cancel out when you multiply by the conjugate. What you need to do is ensure that you're not dividing by 0. It would help to show your work when you do that.

 

[OmniNewton]

You don't necessarily need terms to cancel out when you multiply by the conjugate. What you need to do is ensure that you're not dividing by 0. It would help to show your work when you do that.

OK I will post a screen shot of my work

 

[OmniNewton]

OK I will post a screen shot of my work

 

[axmls]

So, notice what's causing difficulties. You want to make sure you don't divide by 0, but multiplying by the conjugate of the numerator won't take away that term in the denominator that goes to 0. So perhaps try multiplying by the conjugate of the bottom.

 

[SteamKing]

My bad the problem is actually h----> 1 I have been looking at the problem for 2 hours :p

Sometimes, a limit can be evaluated by simply plugging in the limit value, as suggested by axmls. It's only when you get either an infinite or otherwise undefined result by substitution that you need to resort to more subtle manipulations.

 

[OmniNewton]

Hi, I tried taking the conjugate of the bottom and a similar problem has occurred:

The problem is still of the form (0/0) when substituted in

 

[OmniNewton]

Sometimes, a limit can be evaluated by simply plugging in the limit value, as suggested by axmls. It's only when you get either an infinite or otherwise undefined result by substitution that you need to resort to more subtle manipulations.

How can you not divide by zero if the zero term is not eliminated?

 

[SteamKing]

How can you not divide by zero if the zero term is not eliminated?

Well, this problem keeps changing, for some reason.

 

[OmniNewton]

Well, this problem keeps changing, for some reason.

Sorry it should have only been that one change to make it so h---> 1 instead of h---> 0

 

[Ray Vickson]

OK I will post a screen shot of my work

No, no, please don't. Screen shots are discouraged; typing it all out is much preferred, because the results are essentially device-independent.

 

[OmniNewton]

No, no, please don't. Screen shots are discouraged; typing it all out is much preferred, because the results are essentially device-independent.

I simplified the limit to the following state:

 

[OmniNewton]

The following limit should actually simplify to this:

 

[OmniNewton]

I figured out the solution! One must multiply the equation by both conjugates!

Thank you for the Guidance.

 

[SammyS]

I figured out the solution! One must multiply the equation by both conjugates!

Thank you for the Guidance.

Right, in that you need to use both the conjugates, the one for the numerator as well as the one for the denominator.

By the way, it's not a equation. An equation has two sides separated by an equal sign.