💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Determine the infinite limit.$\displaystyle \lim_{x \to 2^+}\frac{x^2 - 2x -8}{x^2 -5x +6}$

## $\lim _{x \rightarrow 2+} \frac{x^{2}-2 x-8}{x^{2}-5 x+6}=\lim _{x \rightarrow 2+} \frac{(x-4)(x+2)}{(x-3)(x-2)}=\infty$ since the numerator is negative and the denominator approaches 0 throughnegative values as $x \rightarrow 2^{+}$

Limits

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##### Top Calculus 1 / AB Educators  ##### Heather Z.

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in this problem, we want to find a limit of this expression as X approaches to from the positive side, it's going to be an infinite limit. So that means this expression is either going to approach positive infinity or negative infinity. As X approaches to from the positive side. Uh And the way we're going to ah approach this problem is we are going to factor uh the numerator and the denominator of this rational expression. So x squared minus two x minus eight will factor into x minus four Times X-plus two And then the denominator will factor into X -3 Times X -2. Now, as X approaches to from the positive side, x minus four will approach to -4 or negative too, X plus two will approach for as X approaches to from the positive side, As X is approaching two, X-plus two will be approaching for the x minus three factor in the denominator as X approaches to from the positive side. Well, as X is approaching two to minus three negative one, X minus three will approach negative one as X approaches to from the positive side. So these three factors all approach uh an actual number as X approaches to from the positive side. Uh this factor Um this X -2 factor, this is the one we need to look at closely, What does this X -2 factor approach as X approaches to from the positive side? Well, if x is approaching two from the positive side, that means we're getting near to but we're always a little bit higher than to like, you know, 2.42 point 32.22 point 12 point oh 12.1. We keep getting closer to two but from the positive side always a little bit above two. So if X is approaching two from the positive side, the closer you get to two from the positive side, X -2 is going to be a tiny positive number. Okay, so this gets closer and closer to zero. Okay, this approaches zero are from you know, the positive site. So as x approaches to from the positive side, X -2 is going to be a very small positive number. Give you a little example. I suppose X is approaching two from the positive side, we're really close to two from the positive side. Like 2.001 Then 2.001 -2 is .001. So .001 um is A positive number. .001 is positive but it's really close to zero. So it's a really tiny positive number. Now when you take a really tiny positive number um and times by a negative one. Ok, these two together the x minus three times x minus two as x is approaching two from the positive side X -2 is getting close to zero tiny number but it's still positive but this x minus tree factor was approaching negative one. So this might be 0.1 but this is close to negative one. So negative one times 10.1 is a negative 0.1. So the whole denominator as X approaches to from the positive side okay approaches zero but is negative. So a negative tiny number. So the whole denominator, the x minus three times X minus two approaches zero and it's negative. Okay, so think of a negative tiny number Like -201. Now we already know uh that the whole numerator um the whole numerator is approaching uh negative two times four, which is a positive eight. I'm sorry, a negative eight. So the whole numerator approaches -8. So ultimately as X approaches to from the positive side, the numerator approaches -8. And the denominator uh becomes a tinier and tinier negative number. Well, first of all, a negative eight divided by a tiny negative number. Just the fact that it's a negative tiny number, A negative divided by a negative is going to be positive. So we know the limit is going to you know, be approaching a positive amount. No uh negative eight divided by a tiny negative number. When you take a number like eight and divided by a tiny number. For example eight divided by point oh one. There's going to be 800 Eight divided by a tinier number. Like .0001 is going to be an even larger number 80,000. So negative eight over a tiny negative number. As the tiny negative number gets even tinier, uh This entire limit is going to approach positive infinity. So the limit of dysfunction as X approaches to from the positive side is positive infinity. Last but not least, I wanted to show you graphically what this function looks like and to support the answer that we found for our limit. Um you can see that as X approaches to from the positive side, this function is heading towards positive infinity. Temple University

#### Topics

Limits

Derivatives

##### Top Calculus 1 / AB Educators  ##### Heather Z.

Oregon State University  ##### Michael J.

Idaho State University

Lectures

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