Question

find the limit as x approaches 0

f(x)= [(e^x)+x]^(1/x), answer should be e^2 but how do I get there?

Answer 1

i think it had something to do with natural log

Answer 2

of both sides

Answer 3

well, I didn't get the idea here to reach the answer you posted, are you sure of the whole notations

Answer 4

[e to the x plus x] all to the power of 1/x

Answer 5

OKay I might be able to get this..I am posting the solution

Answer 6

ok

Answer 7

\[\lim_{x \rightarrow 0 } (e^x + x + 1 - 1)^{\frac{1}{x}}\]
\[\lim_{x \rightarrow 0 } (1+ ( e^x + x - 1))^{\frac{1}{x}}\]
\[e^{\large{\lim_{x \rightarrow 0}(e^x + x -1)*\frac{1}{x}}}\]

Answer 8

\[e^{\large{\lim_{x \rightarrow 0}\frac{e^x -1}{x} +1 }}\]

Answer 9

Okay If I try L'Hospital than it might work

Answer 10

\[e^{\large{\lim_{x \rightarrow 0}e^x+1}}\]

Answer 11

Now \(x \rightarrow 0 \text{ that implies } e^x \rightarrow 1\)

Answer 12

Which Makes it \(e^2\)

Answer 13

Yay I got it ...hmm I hope I did that in right way seems right to me I didn't violated any Mathematical Principle ah well you can check ...I gotta go Now ... Good Luck (y)

Answer 14

good

Answer 15

\[\lim_{x \rightarrow 0}(e^x+x)^{\frac{1}{x}}\] best bet may be to get the variable out of the exponent via the log, so first find
\[\lim_{x\rightarrow 0}\frac{1}{x}\ln(e^x+x)\] which has form 0/0
l'hopital turns this into
\[\lim_{x\rightarrow 0}\frac{e^x+1}{e^x+x}=2\] by substitution. therefore the limit of the log is 2, so the limit of the original expression is
\[e^2\]