Question

please help me .. lim x-> infinity (x-squareroot x ssquare minus 3x )

Answer 1

\[\lim_{x\to\infty}\left(x-\sqrt{x^2-3x}\right)\cdot\frac{x+\sqrt{x^2-3x}}{x+\sqrt{x^2-3x}}\]
\[\lim_{x\to\infty}\frac{x^2-(x^2-3x)}{x+\sqrt{x^2-3x}}\]
\[\lim_{x\to\infty}\frac{3x}{x+\sqrt{x^2-3x}}\]
In the denominator, factor a \(\sqrt{x^2}\), so that \(\sqrt{x^2-3x}=\sqrt{x^2}\sqrt{1-\dfrac{3}{x}}\).
\[\lim_{x\to\infty}\frac{3x}{x+\sqrt{x^2}\sqrt{1-\frac{3}{x}}}\]
\(\sqrt{x^2}=|x|=x\), since \(x\to\infty\) (\(x\) is positive, so \(|x|=x\)).
\[\lim_{x\to\infty}\frac{3x}{x+x\sqrt{1-\frac{3}{x}}}\]

Answer 2

oh . so the last answer is ?

Answer 3

thanks a lot ..

Answer 4

yw

Answer 5

this is about l'hospital rule ..

Answer 6

So you have to use it? I don't see why, since you get the answer right away without it.

Answer 7

this is my assignment frm my lecturer ..

Answer 8

At this point
\[\lim_{x\to\infty}\frac{3x}{x+\sqrt{x^2-3x}}\]
you get the indeterminate form \(\dfrac{0}{0}\) so you could apply the rule then. You'll get something that will have to be dealt with in a similar way, where you factor out \(\sqrt{x^2}\) and so on.

Answer 9

oh , alright , thanks ..