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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?

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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*}

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  • $\begingroup$ That is the solution, thanks. $\endgroup$ – Garret Raziel Mar 3 '11 at 22:10
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$$ 3 (\sqrt{x}\sqrt{x-3} -x +2 ) = 3 (\sqrt{x^2-3 x} -x +2 ) = 3 [x (\sqrt{1 -3/x} -1) +2 ]$$ $$=3 [x (1 -3/2x + O(x^{-2}) -1) +2 ] = 3/2 + O(x^{-1})$$ which implies that the limit is $3/2$.

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    $\begingroup$ Not sure if OP will understand this, +1 anyway! $\endgroup$ – Aryabhata Mar 3 '11 at 21:58
  • $\begingroup$ +1 and two comments: first I think $O(x^{-1})$ is better than $O(x)^{-1}. Secondly, in the above we say that the first expression is of *the class* $3/2 + O(x^{-1})$ which imply the limit - we can not take the limit of the class (it might be confusing). $\endgroup$ – AD. May 16 '11 at 7:02
  • $\begingroup$ @AD: thank you for the comments. I edited the answer. $\endgroup$ – Fabian May 17 '11 at 6:59
  • $\begingroup$ Thanks! Just noticed I forgot to press +1. $\endgroup$ – AD. May 17 '11 at 11:50
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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.

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  • $\begingroup$ you mean $f(x) = \sqrt{1-3x}+2x$ right? $\endgroup$ – Please Delete Account Mar 4 '11 at 4:26
  • $\begingroup$ @App: Either works since they differ by 1 so have the same derivative. $\endgroup$ – Bill Dubuque Mar 4 '11 at 4:43
  • $\begingroup$ yes, my mistake. $\endgroup$ – Please Delete Account Mar 4 '11 at 4:46
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Hint:

$$(\sqrt{x} \sqrt{x-3} - x)(\sqrt{x} \sqrt{x-3} + x) = -3x$$

$3/2$ is correct.

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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 $$6-\frac{9}{2}=\frac{3}{2}.$$

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Hint: $$ \begin{align} \lim_{x\to\infty}3\left(\sqrt{x\vphantom{3}}\sqrt{x-3}-x+2\right) &=6+3\lim_{x\to\infty}\left(\sqrt{x\vphantom{3}}\sqrt{x-3}-x\right)\\ &=6+3\lim_{x\to\infty}\frac{-3x}{\sqrt{x\vphantom{3}}\sqrt{x-3}+x}\\ &=6-9\lim_{x\to\infty}\frac1{\sqrt{1-\frac3x}+1} \end{align} $$

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