Evaluate the limit.
lim x approaches 0-
x sqrt. (4 + x^−2)

Question

Evaluate the limit.
lim x approaches 0-
x sqrt. (4 + x^−2)

dumbcow · Answer

$$x \sqrt{4+\frac{1}{x^{2}}} = x \sqrt{\frac{4x^{2} +1}{x^{2}}} = \sqrt{4x^{2} +1}$$

anonymous · Answer

@dumbcow  What?

dumbcow · Answer

i simplified the expression so that limit could be evaluated at x=0

dumbcow · Answer

in its original form if you plug in 0 you get 0*infty which is indeterminate

anonymous · Answer

Good for you.

anonymous · Answer

so the answer is 1?

anonymous · Answer

but it's telling me it's not 1 :(

anonymous · Answer

$$\frac{x}{|x|} \sqrt{4x^2+1}$$

dumbcow · Answer

http://www.wolframalpha.com/input/?i=lim+x*sqrt%284%2Bx^-2%29+as+x-%3E0 ahh sorry i forgot to include neg since we are approaching from left

anonymous · Answer

but since it's factor shouldn't it be 1 regardless since it's x^2

muckushla: why the absolute bar?

anonymous · Answer

If you factor the 1/x^2 out rather than taking the x into the radical, you get the abs value in the denominator like @mukushla the result is that the limit from the left, does not equal the limit from the right and the limit does not exist... "dne"

anonymous · Answer

How to tell which way to go is the hard part.  I see both ways working, but graphing on a calculator shows the dne result.

anonymous · Answer

I guess that taking the x into the radical loses information, whereas factoring out with the absolute value retains information.

anonymous · Answer

oh i see thank you