Question

I will write problem in comment. I need some hints. :)

Answer 1

I can't write equation for some reason, I hope you will understand this
lim x to 0+ (e^(5x)+x^2)^(1/x)

Answer 2

Have you tried graphing it?

Answer 3

no

Answer 4

Are you against that method of solving it?

Answer 5

\[\Large \lim_{x\to 0^+} (e^{5x}+x^2)^{\frac{1}{x}}\]
Is this your question?

Answer 6

@gorica ?

Answer 7

Yes

Answer 8

@klik I am not allowed to solve it by graphing

Answer 9

Ok let's talk about this. I've got a little time.

Answer 10

I'm a little rusty at doing limits, but I think I can help.

Answer 11

Please, write whatever you think would help and I will try/

Answer 12

I've got the answer.

Answer 13

I want to try to prompt the answer from you though, instead of just giving it to you. I think it will help you learn. Side note, is there any way to make this send a message by pressing enter?

Answer 14

Take the ln and use L'Hospital rule you find that the limit is e^5

Answer 15

I tried to write it in a form lim (1+e^{5x}+x^2-1)^{1/x}=lim (1+a)^[{1/a}*a*{1/x}], where a=e^{5x}+x^2-1. Sorry for writing equations like that, but I can't write equations today, I don't know what's the problem with it :(

Answer 16

how to use ln?

Answer 17

Elisa, I think you should take a look at the equation Gorica wrote again. I don't believe that's the correct answer, or the correct method in this case.

Answer 18

Hold on a minute. I think I made a mistake reading your equation. Eliassaab is correct.

Answer 19

\[

y= \left(x^2+e^{5 x}\right)^{\frac{1}{x}}\\
\ln y =\frac {\ln\left(x^2+e^{5 x}\right)} {x}\\

\]

Answer 20

Use L'hospital's Rule
\[\Large

\frac {\frac{2 x + 5 e^{5x}}{x^2 + e^{5 x}}


}


{1}=\frac{2 x + 5 e^{5x}}{x^2 + e^{5 x}}

\]
When x goes to zero, the ratio goes to 5
Hence the orginal function goes to e^5

Answer 21

\[
\ln y \to 5\\
y=e^{\ln y} \to e^5

\]