# Video: APCALC04AB-P1A-Q20-140156343204

Find the lim_(π₯ β 0) π₯Β²/(tanΒ² (3π₯)).

02:52

### Video Transcript

Find the limit as π₯ approaches zero of π₯ squared over tan squared three π₯.

The first thing we should always try to do when evaluating limits is use direct substitution. In this case so, we get zero squared over tan squared zero, which is zero over zero. Now thatβs undefined. So we have the indeterminate form. Weβre going to need to find another way then to evaluate our limit.

To achieve this, weβre going to recall the limit of π₯ approaches zero of π of π₯ over tan ππ₯. Once again, if we used direct substitution here, we find that itβs zero over zero, which is undefined. We can use LβHΓ΄pitalβs rule to evaluate this limit. The part weβre interested in says that if the limit as π₯ approaches zero or of π of π₯ over π of π₯ is equal to zero over zero, then this limit is actually equal to the limit as π₯ approaches zero of π prime of π₯ over π prime of π₯.

So in this case, we need to find the limit as π₯ approaches zero of the derivative of ππ₯ over the derivative of tan ππ₯. Well, π is a constant. So the derivative of ππ₯ is π. The derivative of tan of π₯ is sec squared π₯. And in fact, the derivative of tan of ππ₯ is π sec squared ππ₯. We divide through by π. And we find that our limit is equal to one over sec squared ππ₯.

This time, we can perform direct substitution. It becomes one over sec squared zero, which is one over one, which is of course simply equal to one. And this is really useful. We found that the limit as π₯ approaches zero of ππ₯ over tan ππ₯ is equal to one for constant values π. And in fact, this general result can be quoted.

So our next step is to rewrite our limit. Letβs represent π₯ squared over tan squared three π₯ as π₯ over tan three π₯ all squared. And we can further rewrite this as π₯ over tan three π₯ times π₯ over tan three π₯.

We then recall that the limit of a product of two functions is equal to the product of the limits of each of those functions. So we can rewrite this as the limit as π₯ approaches zero of π₯ over tan of three π₯ times the limit as π₯ approaches zero of π₯ over tan of three π₯.

We now go back to the limit we derived earlier. And we notice weβve almost got it in this form. However, our limit is ππ₯ over tan of ππ₯. So weβre going to multiply the numerator and denominator of both of our fractions by three. Remember, this is just equivalent to multiplying by one. And it doesnβt change the size of the fraction. We then recall that the limit of a constant times a function is equal to the constant times the limit of that function. So we could take out a constant of one-third from each of our functions as shown.

And weβre now ready to work out the limit as π₯ approaches zero of three π₯ over tan of three π₯. In each case, itβs simply one as we calculated earlier. And we find that the limit as π₯ approaches zero of π₯ squared over tan squared three π₯ is a third times one times a third times one, which is equal to one-ninth.