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$\lim_{x \to -\infty}\dfrac{\sqrt{10+11x^2}}{12+13x}$

= multiply top and bottom by $\dfrac{1}{x}=-\dfrac{1}{\sqrt{x^2}}$

My question is, why is the negative sign in front so crucial, I don't get it:

$\lim_{x \to -\infty}-\dfrac{\sqrt{10/x^2+11x^2/x^2}}{12/x+13x/x}=-\dfrac{\sqrt{11}}{13}$

Why wouldn't the answer be $\dfrac{\sqrt{11}}{13},$ the limit as $x$ goes to positive infinity, $x \rightarrow \infty$?

Thank you.

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  • $\begingroup$ you dont need the minus signal =) . Work without it =) $\endgroup$ – math student Sep 25 '14 at 4:23
  • $\begingroup$ @mathstudent that's how it was presented in class $\endgroup$ – Emi Matro Sep 25 '14 at 4:24
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    $\begingroup$ when i learned, i dont learned with the minus signal. i will write my solution to help you with something $\endgroup$ – math student Sep 25 '14 at 4:27
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    $\begingroup$ my last comment it was not right .... i will think more about your question $\endgroup$ – math student Sep 25 '14 at 4:29
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Maybe this is easier to visualize: \begin{align*} \lim_{x \to -\infty} \frac{\sqrt{10+11x^2}}{12+13x} & = \lim_{x \to -\infty} \frac{\sqrt{x^2\big(\frac{10}{x^2}+11\big)}}{x\big(\frac{12}{x}+13\big)} \\ & = \lim_{x \to -\infty} \frac{\sqrt{x^2} \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\ & = \lim_{x \to -\infty} \frac{|x| \sqrt{\frac{10}{x^2}+11}}{x\big(\frac{12}{x}+13\big)} \\ & = \lim_{x \to -\infty} -\frac{\sqrt{\frac{10}{x^2}+11}}{\frac{12}{x}+13} \quad\text{because $\frac{|x|}{x}=-1$ when $x < 0$}\\ & =-\frac{\sqrt{11}}{13} \end{align*} But intuitively, if the limit exists, then it must be negative because the numerator is always positive while the denominator is negative whenever $x < -1$.

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Set $-\dfrac1x=h$

$F=\lim_{x \to -\infty}\dfrac{\sqrt{10+11x^2}}{12+13x}$

$=\lim_{h\to0^+}\dfrac{\sqrt{10h^2+11}}{|h|}/\dfrac{12h-13}h$

As $h\to0,h\ne0$ and $|h|=h$ as $h>0$

$\implies F=\lim_{h\to0^+}\dfrac{\sqrt{10h^2+11}}{12h-13}=\dfrac{\sqrt{11}}{-13}$

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