Question

can anyone solve it with all the processes please:
lim-->2 x−√x+2/√4x+1−3

Answer 1

I might be able to help you with this.

Answer 2

It would help if you could draw your equation. Because there are some parenthesis that might be missing.

Answer 3

You still there?

Answer 4

use L'Hopital rule...

Answer 5

\[\lim_{x \rightarrow 2}\frac{ x -\sqrt{x +2} }{ \sqrt{4x +1}-3 }\]Is this how the question supposed to look, @kabulis ?

Answer 6

yes this is what my question is..

Answer 7

If it is, rationalize both the numerator and the denominator.

Example: Find\[\lim_{x \rightarrow 0}\frac{ \sqrt{x +1}-\sqrt{2x +1} }{ \sqrt{3x +4}-\sqrt{2x +4} }\]
\[=\lim_{x \rightarrow 0}\left( \frac{ \sqrt{x +1}-\sqrt{2x +1} }{ \sqrt{3x +4}-\sqrt{2x +4} } \right)\left( \frac{ \sqrt{x +1}+\sqrt{2x +1} }{ \sqrt{3x +4}+\sqrt{2x +4} } \right)\left( \frac{ \sqrt{3x +4}+\sqrt{2x +4} }{ \sqrt{x +1}+\sqrt{2x +1} } \right)\]
\[=\lim_{x \rightarrow 0}\left( \frac{ (x +1)-(2x +1) }{ (3x +4)-(2x +4) } \right)\left( \frac{ \sqrt{3x +4}+\sqrt{2x +4} }{ \sqrt{x +1}+\sqrt{2x +1} } \right)\]
\[=\lim_{x \rightarrow 0}\left( \frac{ -x }{ x } \right)\left( \frac{ \sqrt{3x +4}+\sqrt{2x +4} }{ \sqrt{x +1}+\sqrt{2x +1} } \right)\]
\[=-\lim_{x \rightarrow 0}\frac{ \sqrt{3x +4}+\sqrt{2x +4} }{ \sqrt{x +1}+\sqrt{2x +1} }\]
\[=(-1)\left( \frac{ \sqrt{3(0)+4}+\sqrt{2(0)+4} }{ \sqrt{0+1}+\sqrt{2(0)+1} } \right)\]
\[=(-1)\left( \frac{ 2+2 }{ 1+1 } \right)\]
\[=-2\]