natural logarithmic functions-solve using limits: limx->2-(e^(3/2-x))

[cegraham123]

limx->2-(e^(3/2-x))

using chain rule and u sub:
g(x)=3/(2-x) f(u)=e^u

limx->2- (3/2-x)=infitnity

so with the common limit:
lim u->infintiy (e^u)=infinity

it is telling me the answer infinity is incorrect, but it is webassign. am i doing something incorrectly?! help!

 

[tkhunny]

limx->2-(e^(3/2-x))

using chain rule and u sub:
g(x)=3/(2-x) f(u)=e^u

limx->2- (3/2-x)=infitnity

so with the common limit:
lim u->infintiy (e^u)=infinity

it is telling me the answer infinity is incorrect, but it is webassign. am i doing something incorrectly?! help!

1) Please be more careful. Once, you wrote 3/(2-x) and twice you wrote 3/2-x. They are NOT the same thing.
2) limx->2- (3/2-x) -- No idea how you magically eliminated your exponential.

Give it another go.

 

[Dr.Peterson]

limx->2-(e^(3/2-x))

using chain rule and u sub:
g(x)=3/(2-x) f(u)=e^u

limx->2- (3/2-x)=infitnity

so with the common limit:
lim u->infintiy (e^u)=infinity

it is telling me the answer infinity is incorrect, but it is webassign. am i doing something incorrectly?! help!

What you wrote is unclear in more than just the missing parentheses; I think the negative sign before the parenthesis is meant to be a superscript on the 2, so you want the limit as x approaches 2 from below: \(\displaystyle \displaystyle \lim_{x\rightarrow 2^-}e^{\frac{3}{2-x}}\).

If so, then that will affect your answer. Think about that. Infinity is not always infinity ...

In the future, if you are not sure whether what you type will be read as intended, be sure to explain it in words. It is not necessary to learn to typeset as I did, as long as you say enough about what you write, and use things like parentheses and spaces effectively!