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Calculus/limits question
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Question
I'm stuck on these two questions:
Evaluate (1-lnx)/e^(1/x) as x→0+
and (1/x - 1/((e^x) - 1)) as x→0+
Any help gratefully received! Thanks.
As x->0, ln(x)->-infinity,so 1-ln(x)->infinity.
As x->0, 1/x->infinity, so e^(1/x)->e^infinity, which is infinity.
This means you can differentiate the top and bottom.
The derivative of 1-ln(x) is 1/x.
The derivative of e^(1/x) is (-1/x²)e^(1/x).
This means this limit is the same as lim(x->0)(1/x)/)(-e^(1/x)/x²).
Note that this is the same as lim(x->0)(x/e^(1/x)).
This goes to 0/infinity, which is 0, so the original problem goes to 0.
As x->0, 1/x-> infinity and -1/(e^x - 1) goes to -1/(1-1), which is -1/0 -> infinity.
Combine the fractions to (e^x - x)/(xe^x). Take derivatives of top and bottom.
We get (e^x - 1)/(xe^x + e^x). Multiplying this by e^(-x)/e^(-x), we get
(1 - e^(-x))/(x+1).
Now if we let x->0, we get (1-1)/1 = 0/1 = 0.
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