Question

lim h of 0 square root of 3+h - square root of 3 all over h . I have to show the steps. I have them but want to make sure ive got them correct.

Answer 1

multiply top and bottom by the conjugate of the numerator. then cancel. then take the limit

Answer 2

my answer is 1

Answer 3

your answer should be
\[\frac{1}{2\sqrt{3}}\]

Answer 4

you need it written out or is that ok?

Answer 5

\[\frac{\sqrt{3+h}-\sqrt{3}}{h}\times \frac{\sqrt{3+h}+\sqrt{3}}{\sqrt{3+h}+\sqrt{3}}\]

\[\frac{3+h-3}{h(\sqrt{3+h}+\sqrt{3})}\]
\[\frac{h}{h(\sqrt{3+h}+\sqrt{3})}\]
\[\frac{1}{\sqrt{3+h}+\sqrt{3}}\] then replace h by 0