lim (x --> ∞) [sqrt(1 + x^2)] / [x^2]

Show and explain your work. Thank you! :)

Answers :EDIT
OOPS!
Got this one wrong.
but the added contributors didn't!...


oo/oo so l'Hospital's applies:

x[sqrt(1+x^2)] / 2x as x-> oo this is still oo/oo so administer again.

After accomplishing so numerator tends to oo and denominator tends to 2. So the absolute as x -> oo is oo/2.

This is undefined, diverges, increases after bound, or tends to beyond depending aloft how you are accomplished it.

Answers :The acquired of sqrt(1+x^2)= x/sqrt(1+x^2). The acquired of x^2 = 2x. Hence your absolute becomes:

lim(x->oo) x/sqrt(1+x^2) / 2x = lim(x->oo) 1/2sqrt(1+x^2)=0

Answers :Hi,

Take the acquired of top and basal separately

Top 1/2(1 + x^2)^(-1/2)*2x = x (1 + x^2)^(-1/2)

Bottom 2x Then put the (1 + x^2) (-1/2) in the denominator and it looks like

x/[2x(1+x^2)^(1/2)]
so we accept 1/[2(1+x^2)^(1/2)] Now as x ---> beyond the denominator increases with out apprenticed so the atom approaches 0.

Hope This Helps

The free-body diagram in the cartoon shows the armament that act on a attenuate rod. The three armament are fatigued to calibration and lie in the even of the screen. Are these armament acceptable to accumulate the rod in equilibrium, or are added armament necessary?

The account looks something like this:

--force1--> _______thin rod________(diagonal up/left force)
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| (force at centermost and due south)
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