Question

An interesting limit, not sure how to properly compute it.

Answer 1

\[ \lim_{x \rightarrow \infty} \frac{x}{\sqrt{x^2-1}}\]

Answer 2

The reason it's weird is when you do L'Hopital's rule it sort of flip-flops around.

Answer 3

rewrite x = sqrt(x^2), since x > 0

Answer 4

or that same function could be written as

Answer 5

and since x->infinity
1/x^2 tends to 0

Answer 6

Another neat approach would be to see that: \[\large \lim_{x \to + \infty} \frac{x^2-1}{x^2} = 1 \implies x^2 -1 \sim x^2 \] So you could make use of that substitution for large values of \(x\).

Answer 7

Ahh ok. My approach was that when I did L'H was since it became the inverse, \[\lim_{x \rightarrow \infty}\frac{x}{\sqrt{x^2-1}}=\lim_{x \rightarrow \infty}\frac{\sqrt{x^2-1}}{x}\]then from that then it seems like the limit has to equal 1 because \[\frac{a}{b}=\frac{b}{a}=1\] for a, b >0

Answer 8

from that approach, limit could equal -1 too....

Answer 9

Yes, but like I said, a,b>0 so it's ok. @hartnn

Answer 10

If you want to make use of L'H, you can try to analyze \(\frac{1}{f}\) rather then \(f\) and then draw your conclusion from there.

Answer 11

oh yeah,
but i am not sure whether you can compare functions inside limits that way...

Answer 12

Since it's x going to infinity and sqrt(x^2-1) to infinity, there's no way it's going to be negative... O.o

Answer 13

yeah, i got how it must be +1