### Video Transcript

Given that the limit as π₯ tends to
negative two of π of π₯ over three π₯ squared is equal to negative three, determine
the limit as π₯ tends to negative two of π of π₯.

In this question, weβ²ve been given
the limit as π₯ tends to negative two of π of π₯ over three π₯ squared. We can break this limit down using
the properties of limits. We have the property for limits of
quotients of functions, which tells us that the limit as π₯ tends to π of π of π₯
over π of π₯ is equal to the limit as π₯ tends to π of π of π₯ over the limit as
π₯ tends to π of π of π₯. For our limit, weβre taking the
limit as π₯ tends to negative two. Therefore, π is equal to negative
two. And we have a quotient of
functions. In the numerator, we have π of
π₯. And in the denominator, we have
three π₯ squared. Applying this rule for limits of
quotients of functions, we obtain that our limit is equal to the limit as π₯ tends
to negative two of π of π₯ over the limit as π₯ tends to negative two of three π₯
squared.

Letβ²s now consider the limit in the
denominator of the fraction. Thatβ²s the limit as π₯ tends to
negative two of three π₯ squared. We can apply direct substitution to
this limit, giving us that the limit as π₯ tends to negative two of three π₯ squared
is equal to three times negative two squared. Negative two squared is equal to
four. We then simplify to obtain that
this limit is equal to 12. We can substitute this value of 12
back in to the denominator of our fraction, giving us that the limit as π₯ tends to
negative two of π of π₯ over three π₯ squared is equal to the limit as π₯ tends to
negative two of π of π₯ all over 12.

However, weβ²ve been given in the
question that the limit as π₯ tends to negative two of π of π₯ over three π₯
squared is equal to negative three. And since this is on the left-hand
side of our equation, we can set our equation equal to negative three. So we now have that the limit as π₯
tends to negative two of π of π₯ over 12 is equal to negative three. We simply multiply both sides of
the equation by 12. Here, we reach our solution which
is that the limit as π₯ tends to negative two of π of π₯ is equal to negative
36.