Question

Which of the following best describes 2-√4-x/4x as the limit approaches zero ?

Answer 1

a. if fails to exist because it unbounded
b. it fails to exist because its oscillates
c it exist and equals to 1/16
d. it exist and equals to 2

Answer 2

\[2-\sqrt{4-x/4x}\]

Answer 3

or
\[2-\sqrt{4-x}/4x\]

Answer 4

the second one...

Answer 5

are you sure it isn't this:
\[\lim_{x \rightarrow 0}\frac{2-\sqrt{4-x}}{4x}\]?

Answer 6

\[\lim_{x \rightarrow 0}(2-\frac{\sqrt{4-x}}{4x})\]
if it really is this one...
maybe we can combine the fractions and see what happens if we plug in 0

Answer 7

\[\frac{ 2-\sqrt{4-x} }{ 4x } \times \frac{ 2+\sqrt{4-x} }{ 2+\sqrt{4-x} }\]

Answer 8

sorry i was a way.... the first step is as @nelsonjedi wrote

Answer 9

you don't need to multiply these together just the numerator and if look carefully you would know the answer.... the mid term should cancel out

Answer 10

ido you still need help?

Answer 11

\[\lim_{x\rightarrow 0}(2-\frac{\sqrt{4-x}}{4x})\]\[\lim_{x\rightarrow 0} \left[\frac{2(4x)}{4x}-\frac{\sqrt{4-x}}{4x}\right]\]\[\lim_{x \rightarrow 0} \left[\frac{8x-\sqrt{4-x}}{4x}\right]\]\[\lim_{x\rightarrow 0} \left[ \frac{8x-\sqrt{4-x}}{4x} \cdot \frac{8x +\sqrt{4-x}}{8x +\sqrt{4-x}}\right]\]\[\lim_{x\rightarrow 0}\left[ \frac{16x^2-(4-x)}{4x(8x+\sqrt{4-x})}\right]\]\[\lim_{x \rightarrow 0} \left[\frac{16x^2 +x -4}{4x(8x+\sqrt{4-x})}\right]\]think you can simplify from here on out?