Question

Another problem in Limits :

\(\lim _ { x\rightarrow a} \cfrac{\sqrt{a+2x} - \sqrt{3x} }{\sqrt{3a + x} -2\sqrt{x}}\)

Answer 1

Should I apply LH rule here? @eliassaab

Answer 2

i did not get your question

Answer 3

@72595210002 - it is a question of limits, I have to evaluate the limit.

Answer 4

I believe multiply top and bottom by conjugate of the numerator and the denominator will eliminate (a-x)

Answer 5

\[
\frac{\text{Num}'(x)}{\text{Den}'(x)}=\frac{\frac{1}{\sqrt{a+2
x}}-\frac{\sqrt{3}}{2 \sqrt{x}}}{\frac{1}{2 \sqrt{3
a+x}}-\frac{1}{\sqrt{x}}}=\\\frac{\sqrt{3 a+x} \left(2 \sqrt{x}-\sqrt{3}
\sqrt{a+2 x}\right)}{\sqrt{a+2 x} \left(\sqrt{x}-2 \sqrt{3 a+x}\right)}\to \frac{2}{3 \sqrt{3}}
\text{ when } x \to a
\]

Answer 6

and the limit is indeed 2/(3 sqrt(3))

Answer 7

Got it... it is too confusing for me .. :( but am trying to get through it.

Answer 8

Thanks again @eliassaab , and @sourwing

Answer 9

YW