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- We have an interesting
problem or exercise here.
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Find a such that the
limit as x approaches zero
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of the square route of four plus x
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minus the square route of four
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minus a times x,
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all of that over x, is equal to 3/4.
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And like always, I encourage you
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to pause the video and give a go at it.
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So assuming you have had your go,
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now let's do this together.
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So when you just try to superficially
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evaluate this limit here,
if x approaches zero,
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so if you're just trying to evaluate
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this one x equals zero,
you're going to get...
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let me just try to evaluate the limit.
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As x approaches zero
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of the square route of four plus x
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minus the square route of four minus ax,
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all of that over x.
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Well, this right over here is going to be
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just the principal root of four,
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because four plus zero is four.
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This right over here is just going to be
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the principal root of four,
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because no matter what a is,
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a times zero is going to be zero,
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so you're going to be
left with four minus zero,
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so it's just the principal route of four.
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So you're going to have two.
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This whole thing is going to be two.
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If you just were to substitute x there,
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so this whole thing is two.
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This whole thing right over here
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is going to be two as well.
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You're going to have two minus two,
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and then as x approaches zero,
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this is going to be zero.
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So this looks like we are going...
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we are getting an indeterminate form.
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And when you get to something like this,
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you start to say, "Well,
L'Hopital's rule might apply."
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If I get zero over zero,
or infinity over infinity,
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well, this limit is going
to be the same thing
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as the limit as x approaches zero.
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This is going to be the same thing
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as the limit as x approaches zero
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of the derivative of the numerator
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over the derivative of the denominator.
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So what is the derivative
of the numerator?
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Actually, let me just do the derivative
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of the denominator first,
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because the derivative
of x, with respect...
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Oh, I may have to do that
in a different color.
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The derivative of x with respect to x
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is just going to be one.
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But now let me take the derivative
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of this business up here.
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The derivative...
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The derivative of this with respect to x.
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So this is four plus x to the 1/2 power.
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So this is, the derivative of this part
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is going to be 1/2 times four
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plus x to the negative 1/2 power.
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And so the derivative of
this part right over here...
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Let's see, here the...
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The chain rule applied
here with a derivative
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of four plus x is just one, so we just
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multiply this thing by one.
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But here the chain rule, the derivative of
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four minus ax, with respect
to x, is negative a.
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Now we multiply that,
and we're going to have
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this negative out front,
so this is going to be
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plus a.
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Plus a times...
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times 1/2
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times four minus ax
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to the negative 1/2 power.
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I just used the power
rule and the chain rule
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to take the derivative here.
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And so what is is this going to be?
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Well, this is going to be equal to...
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This is going to be equal
to something over one.
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So we have up here, as x approaches zero,
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this is going to be,
this part, four plus zero
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is just four to the negative 1/2 power.
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Well, that's just going to be 1/2.
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Four to the 1/2 is two,
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four to the negative 1/2 is 1/2.
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And then as x approaches zero here,
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this is going to be four
to the negative 1/2,
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which is once again 1/2.
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So what does this simplify to?
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We have 1/2 times 1/2, which is 1/4.
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That's that there.
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And then over here I have
a times 1/2 times 1/2,
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so that's going to be plus a over four,
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and so this is the same thing as just
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a plus one over four.
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And we say that this
needs to be equal to 3/4.
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This needs to be equal to 3/4.
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That was our original problem.
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So that needs to be equal to 3/4,
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and now it's pretty straightforward
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to figure out what a needs to be.
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A plus one needs to be equal to three,
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or a is equal to 2.
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And we are done.