Video Transcript
Determine the limit as π₯
approaches zero of nine minus nine cos of seven π₯ over three π₯.
First, we notice that we can cancel
a factor of three from the top and bottom of this fraction, leaving us with this
limit. Next, we notice that we can factor
three out of the limit. If we attempt to do direct
substitution at this point, we will see that we obtain zero over zero, which is
undefined.
Letβs instead use the fact that the
limit as π₯ approaches zero of one minus cos of π₯ over π₯ is equal to zero. We can in fact adapt this formula
in order to find that the limit as π₯ approaches zero of one minus cos of ππ₯ over
π₯ is equal to zero, where π is just a constant. We can do this by substituting ππ₯
in for π₯ into the first rule. We obtain that the limit as ππ₯
goes to zero of one minus cos of ππ₯ over ππ₯ is equal to zero.
Now since π is a constant and ππ₯
is going to zero, this means that π₯ is also going to zero. Instead of writing ππ₯ goes to
zero, we can simply write π₯ goes to zero. Next, we notice we have a factor of
π in the denominator of this fraction. And so we can factorize this π out
of the limit.
Next, we can multiply both sides of
the equation by π. Since the right-hand side is zero,
zero times π gives us zero. So the right-hand side remains as
zero. And so now we have found that the
limit as π₯ goes to zero of one minus cos of ππ₯ over π₯ is zero, which is what we
were trying to show. By substituting π equals seven
into this formula, we can see that our limit here is simply zero. And since three multiplied by zero
is simply zero, we find that the solution here is simply zero.
