Video: Using Limits Property and Direct Substitution to Find the Limit of a Function at a Point

Given that lim_(π‘₯ β†’ βˆ’2) (𝑓(π‘₯))/(3π‘₯Β²) = βˆ’3, determine lim_(π‘₯ β†’ βˆ’2) 𝑓(π‘₯).

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Video Transcript

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

In this question, weβ€²ve been given the limit as π‘₯ tends to negative two of 𝑓 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‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ over the limit as π‘₯ tends to π‘Ž of 𝑔 of π‘₯. For our limit, we’re taking the limit as π‘₯ tends to negative two. Therefore, π‘Ž is equal to negative two. And we have a quotient of functions. In the numerator, we have 𝑓 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 𝑓 of π‘₯ over the limit as π‘₯ tends to negative two of three π‘₯ squared.

Letβ€²s now consider the limit in the denominator of the fraction. Thatβ€²s 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 𝑓 of π‘₯ over three π‘₯ squared is equal to the limit as π‘₯ tends to negative two of 𝑓 of π‘₯ all over 12.

However, weβ€²ve been given in the question that the limit as π‘₯ tends to negative two of 𝑓 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 𝑓 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 𝑓 of π‘₯ is equal to negative 36.

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