Limit of $\lim\limits_{x \to\infty} 3\left(\sqrt{\strut x}\sqrt{\strut x-3}-x+2\right)$ I have to compute this limit:
$$\lim_{x \to\infty} 3(\sqrt{\strut x}\sqrt{\strut x-3}-x+2)$$
wolfram alpha says that answer is $\frac{3}{2}$, but I can't get why. Does anyone know how to get this limit?
 A: $$ 3 (\sqrt{x}\sqrt{x-3} -x +2 ) =
3 (\sqrt{x^2-3 x} -x +2 ) = 3 [x (\sqrt{1 -3/x} -1) +2 ]$$
$$=3 [x (1 -3/2x + O(x^{-2}) -1) +2 ] = 3/2 + O(x^{-1})$$
which implies that the limit is $3/2$.
A: HINT $\ $ It is simply a first derivative: $\: $ changing variables $\rm\ x\to 1/x\ $ transforms it to
$$\rm\displaystyle\ 3\ \lim_{x\to\ 0}\ \frac{\sqrt{1-3x}+2\ x-1}{x}$$
But this $\rm\ lim\ $ has the form of a derivative, namely it is simply
$$\rm\displaystyle\ \lim_{x\to\ 0}\ \frac{f(x)-f(0)}x\ =\ f\:'(0) \quad\ \ \ for\quad\ f(x)\ =\ \  \sqrt{1-3x}+2\ x-1$$
Now it is easy to directly calculate that $\rm\ f\:'(0)\ =\ 1/2\ $ so the original limit $\to 3/2\:.\:$ Note that this solution employs only knowledge of the definition of the derivative and basic rules for calculating derivatives of polynomials and powers. It does not employ related more advanced techniques such (binomial) power series or Taylor series, l'Hôpital's rule, the mean-value theorem, etc.
A: Hint:
$$(\sqrt{x} \sqrt{x-3} - x)(\sqrt{x} \sqrt{x-3} + x) = -3x$$
$3/2$ is correct.
A: The two standard techniques work: multiply and divide by the conjugate, then divide both numerator and denominator by the highest power of $x$.
\begin{align*}
\lim_{x\to\infty}3\left(\sqrt{x}\sqrt{x-3} - x+2\right) &= 3\lim_{x\to\infty}\left(\sqrt{x^2-3x} - (x-2)\right)\\
&= 3\lim_{x\to\infty}\frac{(\sqrt{x^2-3x}-(x-2))(\sqrt{x^2-3x}+(x-2))}{\sqrt{x^2-3x}+ (x-2)} \\
&= 3\lim_{x\to\infty}\frac{(x^2-3x) - (x-2)^2}{\sqrt{x^2-3x}+(x-2)}\\
&= 3\lim_{x\to\infty}\frac{x^2 - 3x - x^2 + 4x - 4}{\sqrt{x^2-3x} + (x-2)}\\
&= 3\lim_{x\to\infty}\frac{x - 4}{\sqrt{x^2-3x}+(x-2)}\\
&= 3\lim_{x\to\infty}\frac{\frac{1}{x}(x-4)}{\frac{1}{x}(\sqrt{x^2-3x}+(x-2))}\\
&= 3\lim_{x\to\infty}\frac {1 - \frac{4}{x}}{\sqrt{1 - \frac{3}{x}} + 1 - \frac{2}{x})}\\
&= 3\left(\frac{1}{\sqrt{1}+1}\right) = \frac{3}{2}.
\end{align*}
A: See this MSE page, your question is similar.  Notice your limit can be rewritten as $$6+3\cdot \lim_{x\rightarrow\infty}\left(\sqrt{x^2-3x}-x\right)$$ but in general we have that 
 $$\lim_{x\rightarrow\infty}\sqrt[n]{x^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0}-x=\frac{a_n}{n}$$ which follows from l'hopitals rule.  Hence the answer to your problem is $$6-\frac{9}{2}=\frac{3}{2}.$$
A: Hint:
$$
\begin{align}
\lim_{x\to\infty}3\left(\sqrt{x\vphantom{3}}\sqrt{x-3}-x+2\right)
&=6+3\lim_{x\to\infty}\left(\sqrt{x\vphantom{3}}\sqrt{x-3}-x\right)\\
&=6+3\lim_{x\to\infty}\frac{-3x}{\sqrt{x\vphantom{3}}\sqrt{x-3}+x}\\
&=6-9\lim_{x\to\infty}\frac1{\sqrt{1-\frac3x}+1}
\end{align}
$$
