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What is the limit as $x \to - \infty$ of $$f(x)=\dfrac{(3x+1)\sqrt{4x^2-1}}{5x^2+1}\text{?}$$

I know for x approaching positive infinity the answer is $6/5$. However do you prove for negative infinity it is $-6/5$.

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When $x$ is large, we can discount lower order terms. So $3x+1 \approx 3x$, $5x^2+1 \approx 5x^2$, $4x^2 - 1 \approx 4x^2$ and so $\sqrt{4x^2-1} \approx \sqrt{4x^2} = 2x$. In total, the ratio is roughly $$ \frac{3x \cdot 2x}{5x^2} = \frac{6}{5}. $$

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    $\begingroup$ +1. But we should be careful when going to $-\infty$: since $\sqrt{4x^2}=|2x|$, we actually have $\frac{6}{5} \text{sign}(x)$ for large $x$. $\endgroup$ – Ian Apr 8 '15 at 2:27
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These kinds of questions behave nicely by factoring out powers of $x$:

$$\frac{(3x+1)\sqrt{4x^2-1}}{5x^2+1}=\frac{(3x+1)|x|}{5x^2+1}\sqrt{4-1/x^2}$$

Now notice that the square root has limit $2$ on its own. Can you work out the limit of what's left? As a hint, if $x$ is positive, as is the case when $x\rightarrow\infty$, then $|x|=x$. On the other hand, if $x$ is negative, as is the case when $x\rightarrow-\infty$, then $|x|=-x$.

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