Find the limit as x-> 3- of: e^[2/(x-3)]

[MarkSA]

Hello,

1) Find the limit of x as it approaches 3 from the left of:
e^(2/[x-3])

I'm studying some review problems and the answer to this one appears to be 0, since 2/[x-3] approaches -infinity and e^(-infinity) equals 0.

However, The person who worked this problem out tried to do it a different way, and i'm wondering why what they did didn't work. They did:
y = e^(2/[x-3])
ln y = ln e^(2/[x-3]
ln y = 2/[x-3]
lim as x->3 from left of ln y = lim as x->3 from left of 2/[x-3]
left side stays at ln y, right side should become: -infinity.. right?
ln y = -infinity
e^(ln y) = e^(-infinity)
y = e^(-infinity)
I'm a little confused where you would go from here. The limits have already been done.. any idea?

 

[Deleted member 4993]

Hello,

1) Find the limit of x as it approaches 3 from the left of:
e^(2/[x-3])

I'm studying some review problems and the answer to this one appears to be 0, since 2/[x-3] approaches -infinity and e^(-infinity) equals 0.

However, The person who worked this problem out tried to do it a different way, and i'm wondering why what they did didn't work. They did:
y = e^(2/[x-3])
ln y = ln e^(2/[x-3]
ln y = 2/[x-3]
lim as x->3 from left of ln y = lim as x->3 from left of 2/[x-3]
left side stays at ln y, right side should become: -infinity.. right?
ln y = -infinity
e^(ln y) = e^(-infinity)
y = e^(-infinity) = 0
I'm a little confused where you would go from here. The limits have already been done.. any idea?

 

[galactus]

\(\displaystyle \lim_{x\to{3^{-}}}e^{\frac{2}{x-3}}=e^{2\lim_{x\to{3^{-}}}\frac{1}{x-3}}\)

Now, look at the limit. As \(\displaystyle x\to{3^{-}}\), we get \(\displaystyle e^{-\infty}=0\).

That may not be a very proper way to express it, but I wanted to show you the jist of it.