Question

Find the limit.

Answer 1

\[\lim_{x \rightarrow 2+} e ^{3/(2-x)}\]

Answer 2

The limit of the function approaches zero. I haven't yet determined how to write a rigorous algebraic proof, but we can tell by a probabilistic argument. As 2-x approaches zero, e is raised to a tremendous number. That's negative. So invert the function, and then you have one over e raised to a nigh-infinite number. One over infinity is zero.

Answer 3

Confirmation by WolframAlpha. ;) It's not always correct, though, so be careful.

Answer 4

I checked my book. It's an odd-numbered question so I have the answer: it's 0. I just need to make sense of what you wrote. If 3/(2-x) wasn't an exponent and this wasn't an exponential function, I could find the limit easily.

Answer 5

I just imagined an x value arbitrarily close to 2 from the positive direction. As I get closer, the value of (2-(2+dx)) approaches -dx, and 3/(-dx) approaches infinity. Because this is an exponent, we can invert it and have 3/dx in the denominator. e raised to nigh infinity is, nigh infinity. One over nigh infinity is really just zero plus some arbitrarily small number.

Answer 6

So basically, the 2-x part is getting closer and closer to 0 but doesn't touch 0. A limit with 3/(value close to 0) is infinity. Since this is 2- a value greater than 2, the 2-x part is negative. And because it's exponential, the limit can be inverted to get rid of the negative. Then, like you said, e to the power of infinity is infinity. But because the limit was inverted, it's now 1/(infinity) which is 0.

I like to go step by step in simple terms but is that the idea?

Answer 7

"2- a value"= 2 minus (a value greater than 2)

Answer 8

Yes.