Question

How would I find this limit?

Answer 1

\[\lim_{x \rightarrow \infty}\frac{ e^\sqrt{x} }{ \sqrt{e^x} }\]

Answer 2

Would it be 0?

Answer 3

Can you type it again? it says math processing error

Answer 4

refresh the browser, and yes it is zero

Answer 5

\[\lim_{x \rightarrow \infty}\frac{ e^\sqrt{x} }{ \sqrt{e^x} } = \lim_{x \rightarrow \infty} e^{\sqrt{x} -x/2} = \lim_{x \rightarrow \infty} e^{-x/2} = 0 \]

Answer 6

you can forget about that "sqrt(x)" in brigntness of that "x"

Answer 7

Can LH rule be applied here?

Answer 8

I tried lhopitals and it didn't look good after the first step but rationals way is muuuuch better.

Answer 9

ofcourse saying "forget about sqrt(x)" is dumb.. but i guess it is fine as it gets u correct answer almost always and you're not really looking for rigor here

Answer 10

you could probably take the logarithm and do lhopitals that way actually

Answer 11

How did you get the \[e^\sqrt{x}^{-x \frac{ x }{ 2 }}\]

Answer 12

Oops I meant e^x - x/2

Answer 13

\[\frac{ e^\sqrt{x} }{ \sqrt{e^x} } \]

write the denominator radical in exponent form \[\frac{ e^\sqrt{x} }{ \sqrt{e^x} } = \frac{ e^\sqrt{x} }{ (e^x)^{1/2} } \]

next apply the exponent formula \((a^m)^n = a^{mn}\) to get
\[\frac{ e^\sqrt{x} }{ e^{x/2} } \]

Answer 14

Okay gotcha

Answer 15

So \[\sqrt{x}-\frac{ x }{ 2 }= \frac{ -x }{ 2 }\]

Answer 16

That makes my head hurt. I don't know how you got from that to -x/2

Answer 17

Keep in mind that we want to know if the given expression approaches to some value as we make x super LARGE

Answer 18

look at this quadratic expression :
\[t - \dfrac{t^2}{2}\]

for large values of \(t\), do you agree the the first term is negligible ?
therefore that expression is almost equal to
\[\approx -\dfrac{t^2}{2}\]

?

Answer 19

Yes

Answer 20

plugin \(t = \sqrt{x}\) above

Answer 21

Okay I think I get it, thanks

Answer 22

The way I did it

Answer 23

that was really helpful

Answer 24

Was... \[\large \lim_{x\rightarrow \infty} \frac{e^{\sqrt{x}}}{\sqrt{e^x}}\implies \lim_{x\rightarrow \infty} \frac{e^{x^{1/2}}}{e^{x/2}} \implies \lim_{x\rightarrow \infty} \frac{e^{\infty^{1/2}}}{e^{\infty/2}}\]So the denominator is reaching infinity a lot faster than the numerator, therefore the limit tends to 0

Answer 25

I didn't know how to solve \[\sqrt{x} - \frac{x}{2}\] :\

Answer 26

Okay, thanks!

Answer 27

that actually looks a lot more nice