### Video Transcript

Consider the graph of π of π₯. Find the limit as π₯ tends to one
of π₯ multiplied by π of π₯ all squared.

Here, weβve been asked to find the
limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared. We can see that we have a power of
a function. And so, we can use our rule for the
limits of powers of functions. It tells us that the limit as π₯
tends to π of π of π₯ to the power of π is equal to the limit as π₯ tends to π
of π of π₯ all to the power of π. In the case of our limit, the value
of π is one and the power which we are raising our function to is two. So π is equal to two. Now, we can apply this rule. It tells us that our limit is equal
to the limit as π₯ tends to one of π₯ multiplied by π of π₯ all squared.

Next, we can in fact simplify the
limit within our square further. We have the rule for the limits of
product of functions. It tells us that the limit as π₯
tends to π of π of π₯ multiplied by π of π₯ is equal to the limit as π₯ tends to
π of π of π₯ multiplied by the limit as π₯ tends to π of π of π₯. Now our product of functions is π₯
multiplied by π of π₯. So when we apply this rule to our
limit, we obtain the limit as π₯ tends to one of π₯ multiplied by the limit as π₯
tends to one of π of π₯ all squared.

Now, letβ²s consider the limit as π₯
tends to one of π₯. We can apply direct substitution to
this limit. And we obtain that it is equal to
one. So we can substitute this back in
for our limit, giving us that the limit as π₯ tends to one of π₯ timesed by π of π₯
squared is equal to the limit as π₯ tends to one of π of π₯ all squared. Now all we need to do is find the
limit as π₯ tends to one of π of π₯. In order to do this, we need to use
our graph. We need to find the value of π of
π₯ when π₯ is equal to one. We see that when π₯ is equal to
one, π of π₯ is equal to three. And the graph of π of π₯ near to
the π₯-value of one is a straight line. Therefore, the right limit of π of
π₯ and the left limit of π of π₯ will both agree that this limit is equal to
three. So we can substitute in three for
the limit as π₯ tends to one of π of π₯. So we find that our limit is equal
to three squared. And we can square the three to
obtain our solution that the limit as π₯ tends to one of π₯ multiplied by π of π₯
squared is equal to nine.