Question

lim x->∞ sin(e^x) x ln(x^3+2)+5 / √x

Calculate the limit, if it exists.

Answer 1

Assuming from the post it looks like:

\[\large \lim_{x \rightarrow \infty} \frac{\sin(e^x)x\ln(x^3+2)+5}{\sqrt{x}}\]
?

Answer 2

nope, sin(e^x) is separate and is multiplicated with ln(x^3+2)+5/√x

Answer 3

That's what I thought, the additional 'x' would have just made this diverge

Regardless: we are working with

\[\large \lim_{x\rightarrow \infty}\frac{\sin(e^x)\ln(x^3+2)+5}{\sqrt{x}}\]

What would happen right away if you plugged in a HUGE number here for 'x'? What would the numerator equal? What about the denominator?

Answer 4

Yea, of course, now I see it :)

Answer 5

Cool :D yeah that's usually the first thing to do is to check what would happen if you just plugged it in

Answer 6

The denominator is correct

Answer 7

So wait, are you good? Or still need help?

Answer 8

The problem is still not solved so I´ll be needing some help, yes :)

Answer 9

Wait a minute. I will help you when my guests leave

Answer 10

Thanks! :)

Answer 11

Let me do some steps that will make the problem easier
\[
\lim_{x->\infty} \frac {\ln(x)}{\sqrt x}=0\]

Answer 12

That is easy to prove using L'Hospital's rulle

Answer 13

Near infinity
\[
\ln(x^3+3)\approx \ln(x^3)=3 \ln (x)\]

Answer 14

Combining the above two steps, one has
\[
\frac{\ln(x^3+3)}{\sqrt x} \approx= \frac{3 \ln x}{\sqrt x}\to 0
\]
when \( x\to \infty \)

Answer 15

Now
\[

\left | \frac{ \sin(e^x) \ln(x^3 +2)}{\sqrt x} \right|\le \left |\frac{ \ln(x^3 +2)}{\sqrt x}\right |->0\]

By the steps before

Answer 16

Your limit is zero since the remaining term \( \frac {5}{\sqrt x}\to 0 \) when x goes to infinity

Answer 17

Putting all the steps your original limit will go to zero

Answer 18

Ok, yet another limit is zero? Well, that´s math....

Thank you so much for the help! :)

Answer 19

You are welcome