Video: Evaluating a Given Limit

Consider the graph of 𝑓(π‘₯). Find lim_(π‘₯ β†’ 1) ((π‘₯ β‹… 𝑓(π‘₯))Β²).

02:19

Video Transcript

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

Here, we’ve been asked to find the limit as π‘₯ tends to one of π‘₯ multiplied by 𝑓 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‘Ž of 𝑓 of π‘₯ to the power of 𝑛 is equal to the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ all to the power of 𝑛. In the case of our limit, the value of π‘Ž is one and the power which we are raising our function to is two. So 𝑛 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 𝑓 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‘Ž of 𝑓 of π‘₯ multiplied by 𝑔 of π‘₯ is equal to the limit as π‘₯ tends to π‘Ž of 𝑓 of π‘₯ multiplied by the limit as π‘₯ tends to π‘Ž of 𝑔 of π‘₯. Now our product of functions is π‘₯ multiplied by 𝑓 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 𝑓 of π‘₯ all squared.

Now, letβ€²s 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 𝑓 of π‘₯ squared is equal to the limit as π‘₯ tends to one of 𝑓 of π‘₯ all squared. Now all we need to do is find the limit as π‘₯ tends to one of 𝑓 of π‘₯. In order to do this, we need to use our graph. We need to find the value of 𝑓 of π‘₯ when π‘₯ is equal to one. We see that when π‘₯ is equal to one, 𝑓 of π‘₯ is equal to three. And the graph of 𝑓 of π‘₯ near to the π‘₯-value of one is a straight line. Therefore, the right limit of 𝑓 of π‘₯ and the left limit of 𝑓 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 𝑓 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 𝑓 of π‘₯ squared is equal to nine.

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