Problem

Make a conjecture about the limit by graphing the…

01:24

Problem 48 Hard Difficulty

(a) Find the error in the following calculation:
$$\lim _{x \rightarrow 2} \frac{e^{3 x^{2}-12 x+12}}{x^{4}-16}=\lim _{x \rightarrow 2} \frac{(6 x-12) e^{3 x^{2}-12 x+12}}{4 x^{3}}=0$$
(b) Find the correct limit.

Answer

a. applying L'Hospitals rule to this function is incorrect. This is the error.
b. $\lim _{x \rightarrow 2} \frac{e^{3 x^{2}-12 x+12}}{x^{4}-16}=\frac{e^{0}}{0}=\frac{1}{0}=\infty$

Related Courses

Calculus 1 / AB

Calculus Early Transcendentals

Chapter 3

TOPICS IN DIFFERENTIATION

Section 6

L Hopital's Rule; Indeterminate Forms

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Video Transcript

So in this problem, we're taking a look at a limit, and we're looking at where a mistake was made. So in particular, we're looking at the limit as X goes to two of e race to the three X squared minus 12 X plus 12. And this is divided by X to the fourth minus 16. Um, and then the next step in the problem. They use low bottles rule, which is kind of weird, because you need to have an indeterminant form to be able to use low petals rule. Let's and we we probably and we do not actually have and indeterminant form. Um, so let's go ahead and see that. So when we try to plug in to here in the in the numerator, we end up getting e toothy. Let's see, we have 12 minus 24 plus 12. So this is e to the zero, which is equal toe one, and then in the denominator, we end up with zero. So, yeah, we do not have an indeterminant form, and that's the problem. That's how mistake was made not and in determine INTs form, so I can't use lo petals rule for part B Well, um, let's go ahead and use some of the work that we used that we did in part A. So we have this limit, and let's consider the limits as X approaches to from the right hand side. Um, rather than rewrite this whole expression, let me just draw squiggle there. Now, notice that the numerator of the squiggle tends to one, and the denominator of the squiggle tends to zero from the positive side. So we have positive infinity. Okay? And now let's take a look at the limit as X approaches to from the left hand side of this squiggle expression, the numerator turns to want us before. But now the denominator tends to zero from the negative side. So we actually have negative infinity. So, um, this is a problem. We have that the limits of this whole expression as X approaches to just does not exist. Okay, and that's it

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