Uploaded by TheIntegralCALC on 03.09.2010

Transcript:

Limits - Conjugate Method Example 1

Hi, everyone!

Welcome back to integralcalc.com.

We’re going to be doing another limit problem today.

This one is the limit as h approaches zero of the square root of four plus h minus two over h.

As with any limit problem, the first thing that you should do is try plugging in this number to the function to see if you can just solve the limit out right.

Of course, when you try plugging it in, you get zero on the bottom of the fraction which is not allowed in math at all.

So, we can’t simply plug in zero for h.

The next thing that we should try is factoring.

If we try to factor, we can quickly see there isn’t really a whole lot to factor in either the numerator or the denominator

so that doesn’t seem like it’s going to work very well.

So the third thing that we should try is multiplying by the opposite of the numerator here.

There is really no opposite here to h. What I mean by opposite is the square root of four plus h plus two...

So if we do that... if we multiply by...

Well, I'll show you how... If we multiply this four plus h and then we have minus two over h,

If we multiply it by the opposite of the numerator, and we do that in both the numerator and denominator

plus two... and then the same thing here... four plus h plus two... It's the opposite here of the numerator...

But this fraction reduces to one over one or just one so we’re not actually changing the problem at all because essentially, we’re just multiplying this by one.

What this will allow us to do is simplify and get the h out of the denominator.

So this is what we multiply by, the opposite and when we do that,

we will get the square root of four plus h times the square root of four plus h

which, we'll take away the square root sign and just give us four plus h.

then we have plus two times the square root of four plus h minus two times the square root of four plus h minus four.

This negative two times two here.

And then in the denominator, we'll get h times the square root of four plus h plus two.

So now that we’ve done that, we can go ahead and simplify.

You can see here that these two middle terms will cancel because we have plus and a minus the same thing.

so we’ll end up on the top with four plus h minus four over h times the square root of four plus h plus two.

We have four minus four on the top here so we'll just end up with h over h times the square root of four plus h plus two.

Now we have h in the numerator and h in the denominator and they will cancel.

So we will be left with one over the square root of four plus h plus two.

This is now something that we can plug in zero for h and solve it.

So we’ll end up with zero over the square root of four plus zero which is just four plus two.

The square root of four is two so one over two plus two.

which simplifies to one-fourth.

And that is our final answer.

Hope that helped. See you, guys next time. Bye.

Hi, everyone!

Welcome back to integralcalc.com.

We’re going to be doing another limit problem today.

This one is the limit as h approaches zero of the square root of four plus h minus two over h.

As with any limit problem, the first thing that you should do is try plugging in this number to the function to see if you can just solve the limit out right.

Of course, when you try plugging it in, you get zero on the bottom of the fraction which is not allowed in math at all.

So, we can’t simply plug in zero for h.

The next thing that we should try is factoring.

If we try to factor, we can quickly see there isn’t really a whole lot to factor in either the numerator or the denominator

so that doesn’t seem like it’s going to work very well.

So the third thing that we should try is multiplying by the opposite of the numerator here.

There is really no opposite here to h. What I mean by opposite is the square root of four plus h plus two...

So if we do that... if we multiply by...

Well, I'll show you how... If we multiply this four plus h and then we have minus two over h,

If we multiply it by the opposite of the numerator, and we do that in both the numerator and denominator

plus two... and then the same thing here... four plus h plus two... It's the opposite here of the numerator...

But this fraction reduces to one over one or just one so we’re not actually changing the problem at all because essentially, we’re just multiplying this by one.

What this will allow us to do is simplify and get the h out of the denominator.

So this is what we multiply by, the opposite and when we do that,

we will get the square root of four plus h times the square root of four plus h

which, we'll take away the square root sign and just give us four plus h.

then we have plus two times the square root of four plus h minus two times the square root of four plus h minus four.

This negative two times two here.

And then in the denominator, we'll get h times the square root of four plus h plus two.

So now that we’ve done that, we can go ahead and simplify.

You can see here that these two middle terms will cancel because we have plus and a minus the same thing.

so we’ll end up on the top with four plus h minus four over h times the square root of four plus h plus two.

We have four minus four on the top here so we'll just end up with h over h times the square root of four plus h plus two.

Now we have h in the numerator and h in the denominator and they will cancel.

So we will be left with one over the square root of four plus h plus two.

This is now something that we can plug in zero for h and solve it.

So we’ll end up with zero over the square root of four plus zero which is just four plus two.

The square root of four is two so one over two plus two.

which simplifies to one-fourth.

And that is our final answer.

Hope that helped. See you, guys next time. Bye.