Question

Find the limit:
lim x->2+ e^(3/(2-x))

Answer 1

is it, \(\color{#000000 }{ \displaystyle \lim_{x\to2^+}e^{3/(2-x)} }\) ?

Answer 2

Yes

Answer 3

I will ask you the following question.

\(\color{#000000 }{ \displaystyle \lim_{x\to2^+} \frac{1}{2-x}=? }\)

Answer 4

- infinity?

Answer 5

Yes, that is correct!

Answer 6

\(\color{#000000 }{\LARGE \displaystyle \lim_{x\to2^+}e^{3/(2-x)} =e^{\lim_{x\to2^+}~3/(2-x)} }\)

Answer 7

Well, technically, not, but you can think of that this way

Answer 8

e^(-∞)=1/e^∞=0

Answer 9

But, technically not so right, because we don't raise numbers to infinities.

Answer 10

Okay so my answer would be zero right?

Answer 11

Yes.

Answer 12

\(\color{#000000 }{\large \displaystyle y=\lim_{x\to2^+}e^{3/(2-x)} }\)

\(\color{#000000 }{\large \displaystyle \ln y= \ln\left(\lim_{x\to2^+}e^{3/(2-x)}\right) }\)

\(\color{#000000 }{\large \displaystyle \ln y= \left(\lim_{x\to2^+}\ln e^{3/(2-x)}\right) }\)

\(\color{#000000 }{\large \displaystyle \ln y= \left(\lim_{x\to2^+}{3/(2-x)}\right) }\)

So, if ln(y) tends to negative infinity, then y is 0.

Answer 13

oh okay I see that makes more sense, I knew a ln was needed prove the answer, I just wasn't sure how to

Answer 14

Well, you can just tell that 3/(2-x) will tend to -∞, and thus e^(-∞) = blah blah blah as I showed