Question

Jhghjkkhb

Answer 1

i can do it without using l'hospitals rule but i have to

Answer 2

@satellite73 @surjithayer @jim_thompson5910

Answer 3

so take the derivative of numerator and denominator...

Answer 4

i did and i got \[1/(x/\sqrt(x^2+1))\]

Answer 5

\[\large \lim_{x \to \infty} {x \over \sqrt{x^2+1}}\]

Answer 6

but after you differentiate both dont you have to keep doing it continually because it never stops being an indeterminate?

Answer 7

right ...
\[\large = \lim_{x \to \infty} { \sqrt{x^2+1}\over x}\]

... looks like you're right

Answer 8

why use l'hopital? just asking

Answer 9

the answer is 1 right?

Answer 10

sure

Answer 11

eyeball this one

Answer 12

thats what we are learning right now in class. i know its 1 because i can solve it without l'hopitals

Answer 13

well that is a fine how do you do
if you use l'hopital the first thing you get is
\[\frac{\sqrt{x^2+1}}{x}\] right ?

Answer 14

never mind i read the directions wrong. we're supposed to solve it without l'hopitals and solve a different way. sorry for the trouble

Answer 15

which tells you one thing, if the limit exists and is say \(L\) then \(L=\frac{1}{L}\)