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$$\lim_{x\to -\infty} \frac{x^3}{\sqrt{x^6+4}} $$ When we deal with infinities then 4 is negligible and so the above limit is equal to $$\lim_{x\to -\infty} \frac{x^3}{\sqrt{x^6}} = \frac{x^3}{x^3} = 1$$ but it turns out that this is not equal 1 and it should be -1 , i have no idea why though !

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    $\begingroup$ The bottom is positive always. For negative $x$, the top is negative. Or else recall that $\sqrt{a^2}=|a|$. $\endgroup$ – André Nicolas Jan 19 '15 at 5:32
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    $\begingroup$ @AndréNicolas,what if $\Large\frac{\frac{x^3}{x^3}}{\sqrt {\frac{x^6}{x^6}+\frac{4}{x^6}}}=\frac{1}{\sqrt {1+\frac{4}{x^6}}}=\frac{1}{\sqrt{1+0}}=1$ $\endgroup$ – Vikram Jan 19 '15 at 5:47
  • $\begingroup$ yes exactly @Vikram $\endgroup$ – alkabary Jan 19 '15 at 5:49
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    $\begingroup$ I don't like negative numbers, they are so $\dots$ negative. Let $y=-x$. We want to find the limit as $y$ goes to $\infty$ of $\frac{-y^3}{\sqrt{y^6+4}}$. Easy, little chance of error. $\endgroup$ – André Nicolas Jan 19 '15 at 5:53
  • $\begingroup$ That makes more sense ! @AndréNicolas $\endgroup$ – alkabary Jan 19 '15 at 5:58
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Your process is problematic. In fact, $$\sqrt{x^6}=|x^3|=-x^3$$ as $x\to -\infty$.
Take heed whenever a square root is involved. It will always produce an absolute value, don't leave it out.

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Things become more lucid if we set $-\dfrac1x=h$

$\iff x=-\dfrac1h$

$\implies h\to0^+,\sqrt{h^6}=+h^3$ as $h>0$

$$\lim_{x\to -\infty} \frac{x^3}{\sqrt{x^6+4}} =\lim_{h\to0^+}\dfrac1{-h^3\sqrt{\dfrac{1+4h^6}{h^6}}}$$

$$=\lim_{h\to0^+}\dfrac{\sqrt{h^6}}{-h^3\sqrt{1+4h^6}}=\lim_{h\to0^+}\dfrac{h^3}{-h^3\sqrt{1+4h^6}}=?$$

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  • $\begingroup$ Wish I know the mistake here $\endgroup$ – lab bhattacharjee Jan 19 '15 at 5:51
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    $\begingroup$ While in my opinion it's overly complicated, I can't spot anything wrong with this answer O.o $\endgroup$ – DanZimm Jan 19 '15 at 5:58
  • $\begingroup$ @DanZimm, Not sure about the complication/complexity as setting the limit to $0$ has often eased off my computation $\endgroup$ – lab bhattacharjee Jan 19 '15 at 5:59
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    $\begingroup$ Just an opinion - I find it simpler to compute limits to $\pm \infty$ than anything else! $\endgroup$ – DanZimm Jan 19 '15 at 6:00
  • $\begingroup$ Of course as usual there is nothing wrong with the solution. I also prefer to take troubles to $0$, if the trouble is at $a$, but not so much in the $a=\pm\infty$ case. The sign switch is I think very helpful. $\endgroup$ – André Nicolas Jan 19 '15 at 6:05
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If we think graphically before simplification, we can observe that denominator is always positive, and as $x\to -\infty$ value of $x^3\to -\infty$. So It should be -1.

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