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I'm attempting to evaluate the limit

$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$

I got it reduced to the following

$\lim_{x\rightarrow\infty}\frac{\sqrt{\frac{1}{\left(x-2\right)^{2}}-\frac{3}{\left(x-2\right)^{4}}}+1}{1-\frac{3}{\left(x-2\right)^{2}}-1}$

But putting in $\infty$ I get $\frac{1}{0}$ and, what's worse, Mathematica tells me the limit is equal to $-\infty$. Where am I going wrong?

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  • $\begingroup$ I have amended my solution. I think that solves your doubt. $\endgroup$ – Easy May 7 '13 at 16:03
  • $\begingroup$ This problem looks a little bit unclear to me, if I interpret $x \rightarrow \infty$ as $x \rightarrow + \infty$, I'll have some way to solve it, but if I interpret it as $x \rightarrow - \infty$, then I'll have a completely different way to solve. $\endgroup$ – user49685 May 7 '13 at 18:08
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\begin{align*} &\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}\\ =&\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\ =&\lim_{x-2\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-(x-2)}\\ =&\lim_{z\rightarrow\infty}\frac{1}{\sqrt{z^2-3}-z}\\ =&\lim_{z\rightarrow\infty}\frac{\frac{1}{z}}{\sqrt{1-\frac{3}{z^2}}-1} \end{align*} Note that the absolute value of the last expression is $\infty$ and the denominator is negative as $\sqrt{1-\frac{3}{z^2}}<1$.

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Hint: multiply top and bottom by $\sqrt{x^2-4x+1}+x-2$.

Additional hint upon request: $\lim_{x\rightarrow \infty}\sqrt{x^2-4x+1}=\infty$ and $\lim_{x\rightarrow \infty} x-2=\infty$.

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  • $\begingroup$ You were first :) $\endgroup$ – Mårten W May 7 '13 at 13:58
  • $\begingroup$ @MårtenW, my typing may be fast, but I make a lot of mistakes. $\endgroup$ – vadim123 May 7 '13 at 13:59
  • $\begingroup$ Okay, so I end up with $\lim_{x\rightarrow\infty}\frac{\sqrt{x^{2}-4x+1}+\left(x-2\right)}{-3}$ $\endgroup$ – Dmitri Nesteruk May 7 '13 at 14:22
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$$F=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}$$

$$=\lim_{x\rightarrow\infty}\frac{1}{\sqrt{(x-2)^2-3}-x+2}$$

Let us put $x-2=\sqrt3\csc2\theta$

$x\to\infty\implies \theta\to0$

So, $$F=\lim_{\theta\to0}\frac1{\sqrt3\cot2\theta-(\sqrt3\csc2\theta+2)+2}$$

$$=-\frac1{\sqrt3}\lim_{\theta\to0}\frac{\sin2\theta}{1-\cos2\theta}$$

$$=-\frac1{\sqrt3}\lim_{\theta\to0}\frac{2\sin\theta\cos\theta}{2\sin^2\theta}$$ (using $\sin2\theta=2\sin\theta\cos\theta,\cos2\theta=1-2\sin^2\theta$)

$$=-\frac1{\sqrt3}\lim_{\theta\to0}\cot\theta$$ as $\theta\to0\implies \sin\theta\to0\implies \sin\theta\ne0$

What is $\cot0?$

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    $\begingroup$ This is an epic answer. $\endgroup$ – Dmitri Nesteruk May 7 '13 at 18:51
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Hint: multiply numerator and denominator by the denominator's conjugate:

$$\lim_{x\rightarrow\infty}\frac{1}{\sqrt{x^{2}-4x+1}-x+2}=\frac{\sqrt{x^2-4x+1}+x-2}{x^2-4x+1-x^2+4x-4}=\;\ldots$$

Added under request :

$$\left(\sqrt{x^2-4x+1}+x-2\right)\frac{\frac1x}{\frac1x}=\frac{\sqrt{1-\frac4x+\frac1{x^2}}+1-\frac2x}{\frac1x}\xrightarrow[x\to\infty]{}\ldots$$

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  • $\begingroup$ = $\lim_{x\rightarrow\infty}\frac{\sqrt{x^{2}-4x+1}+\left(x-2\right)}{-3}$, and then? $\endgroup$ – Dmitri Nesteruk May 7 '13 at 14:25
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    $\begingroup$ Check if the added stuff in my answer helps you now, @DmitriNesteruk $\endgroup$ – DonAntonio May 7 '13 at 16:11

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