Question

solve this limit. sqrt(9x+1)/sqrt(x+1)

Answer 1

\[limit \frac{ \sqrt{9x+1} }{ \sqrt{x+1} } as x->infinity\]

Answer 2

\[42\]

Answer 3

l’Hopital’s rule rule wont work

Answer 4

The answer to live the universe and everything.

Answer 5

3.

\[\lim \sqrt{\frac{ 9x+1 }{ x+1}} = \lim \sqrt{\frac{x}{x}\frac{ 9+\frac{1}{x} }{ 1 +\frac{1}{x}}} = \sqrt{\frac{ 9}{1}} = 3\]

Answer 6

man..i would have never figured that out!

Answer 7

its so non-intuative.

Answer 8

so the idea is that the fractions disappear at infinity. what about that x/x. isn't that infintity over infinity? which is not defined

Answer 9

or they just cancel to 1?

Answer 10

x/x will cancel out to one by simple algebra. There are some rules to infinity over infinity that allow it to be defined by how quick they diverge but my teacher didn't go to far into it and I don't really know how to explain it well. And yes, any number over infinity will = 0 as any number is infinitely small compared to infinity.

Answer 11

thank you very much mate! appreciate your help!

Answer 12

wanna try tackle another one?

Answer 13

same kinda concept I think. let me know if your up for it. ill give it a go too

Answer 14

limit as x->0 \[\frac{ \sqrt{x} }{ \sqrt{\sin(x)} }\]

Answer 15

Use l'Hopital's rule, ignore the sqrt.

\[\lim_{x \rightarrow 0} \sqrt{\frac{x}{\sin(x)}} = \sqrt{\lim_{x \rightarrow 0} \frac{x}{\sin(x)}} = \sqrt{\lim_{x \rightarrow 0} \frac{\frac{ d }{ dx }x}{\frac{ d }{ dx }\sin(x)}} = \sqrt{\lim_{x \rightarrow 0} \frac{1}{\cos(x)}} = 1\]

Answer 16

ahh nice. I havent learn't the idea where you can bring the limit inside the square root.

Answer 17

but the questions asks without l'hospitals. "Does l’Hopital’s rule work in the following cases? Explain. Find the limits some
other way"

Answer 18

ok, i figured out the other way. \[limit \sqrt{\frac{ x }{ \sin(x) }} = \sqrt{\frac{ x }{ x }\frac{ 1 }{ \frac{ \sin(x) }{ x } }}\] but we know sin(x)/x = 1 so the answer is 1.

Answer 19

l'hopital's rule requires that you have a function such as f(x)/g(x) where lim f(x) = lim g(x) = 0 or +/-infinity. That is satisfied in this case so it will work.

I am not sure how you would solve this another way. perhaps by making a table and setting x = .1, .01, .001. which would should you the limit.

You cannot pull x our of sin(x) like that.

But yes using the known of sin(x)/x = 1, and that x/sin(x) = (sin(x)/x)^-1 = 1^-1 = 1.

Answer 20

umm. why cant I pull an x like that. doesn't. \[\frac{ x }{ x } \frac{ 1 }{ \sin(x)/x } = \frac{ x }{ \sin(x)}\]

Answer 21

Sorry you could pull x our of sinx like that, I had a mental fart (at work so my attention is divided.)

Answer 22

haha!

Answer 23

hey, what are you studying?

Answer 24

I am going for physics / math. Currently doing my undergrad.

Answer 25

nice, im doing engineering.thats why im on here, asking all the smart guys questions

Answer 26

doing classical mechanics this sem as an elective, and loving it. what year are u in?