Question

If I am given two functions with square roots in them, am I allowed to square the function before I apply l'Hopital's Rule or not?

Answer 1

Give an exampe

Answer 2

For example, in this problem (I already put it in limit form) lim(x->infinity) sqrtx/sqrt(x+1)

Answer 3

Am I allowed to square the whole thing before I l'Hopital?

Answer 4

If I take the square root of my answer anyways

Answer 5

Do you know how to start this?

Answer 6

\[\lim_{x\rightarrow \infty} \frac{\sqrt{x}}{\sqrt{x+1}} \] I would just use LH rule right away. without changing it.

Answer 7

no you can't just square the whole function

Answer 8

you could square the entire function if it was one function.... but here you are dealing with 2 functions.

Answer 9

Why not? My teacher said that we could, but she didn't say anything about before LH

Answer 10

As long as we do the opposite to our answer

Answer 11

\[\lim_{x \rightarrow 3}x=3 \\ \text{ but } \lim_{x \rightarrow 3}x^2=9 \\ 9 \neq 3\]

Answer 12

I'm comparing rates of growth

Answer 13

Freckles, I'm not setting the limit equal to anything

Answer 14

I just have to prove that the limit comes out to a constant other than 0

Answer 15

\[\sqrt{ \lim_{x \rightarrow \infty} \frac{x}{x+1}}\]
you don't need l'hospital

Answer 16

the inside function goes to what ?

Answer 17

Well isn't that the same thing as if I squared the composed functions and took the square root of my answer?

Answer 18

\[\lim_{x\rightarrow \infty} \frac{x^{\color{red}{1}}}{x^{\color{red}{1}}+1}= ~? \]

Answer 19

1

Answer 20

Then you take the square root and get 1

Answer 21

you can't say this:
\[\lim_{x \rightarrow a}f(x) =\lim_{x \rightarrow a}(f(x))^2 \\ \text{ but you can say this } \\ \lim_{x \rightarrow a} f(x) =\lim_{x \rightarrow a} \sqrt{f(x))^2} \text{ provided } f(x)>0\]

Answer 22

No, the square root of the LIMIT

Answer 23

the whole thing

Answer 24

wait

Answer 25

\[\lim_{x \rightarrow \infty}f(x)=\sqrt{\lim_{x \rightarrow \infty} f(x)^{2}}\]

Answer 26

Why can't I do that

Answer 27

I didn't say you couldn't do that

Answer 28

Thats what I meant

Answer 29

That's legit.

Answer 30

oh i thought you just meant square the whole thing and be done with it

Answer 31

lol

Answer 32

I square the composed functions and take the square root of the answer xD

Answer 33

Same, freckles.

Answer 34

Sorry!!!

Answer 35

Yes, and because \(x\rightarrow \infty\) it works.

Answer 36

Ok! So I can square before I use l'hopital?

Answer 37

but I think the problem is easier if we just look at it like this:
\[\lim_{x \rightarrow \infty}\sqrt{\frac{x}{x+1}}=\sqrt{\lim_{x \rightarrow \infty} \frac{x}{x+1}}\]
and just find the limit inside

Answer 38

and yes you can use l'hopistal to find \[\lim_{x \rightarrow \infty}\frac{x}{x+1}\]
but it isn't required

Answer 39

\[\lim_{x \rightarrow \infty}\frac{1}{1+\frac{1}{x}}=\frac{1}{1+0}=\frac{1}{1}=1\]

Answer 40

\[\lim_{x \rightarrow \infty}\sqrt{\frac{x}{x+1}}=\sqrt{\lim_{x \rightarrow \infty} \frac{x}{x+1}}=\sqrt{1}\]

Answer 41

OH OK. Thank you guys so much!!

Answer 42

I didn't know you could take the limit inside a function like that.