Jhghjkkhb

Question

Jhghjkkhb

anonymous · Answer

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

anonymous · Answer

@satellite73 @surjithayer @jim_thompson5910

paxpolaris · Answer

so take the derivative of numerator and denominator...

anonymous · Answer

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

paxpolaris · Answer

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

anonymous · Answer

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

paxpolaris · Answer

right ...
$$\large = \lim_{x \to \infty} { \sqrt{x^2+1}\over x}$$

... looks like you're right

anonymous · Answer

why use l'hopital? just asking

paxpolaris · Answer

the answer is 1 right?

anonymous · Answer

sure

anonymous · Answer

eyeball this one

anonymous · Answer

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

anonymous · Answer

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 ?

anonymous · Answer

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

anonymous · Answer

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