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The problem is evaluate

$$\lim_{x\to\infty}(x^2-\sqrt{x^4 + 7x^2 + 1})$$

I understand all of the calculus involved, but am having trouble figuring out how to get started with the algebra. I have tried factoring and using conjugates, but the only answer I am able to get is $-7$, which is incorrect. Any help would be appreciated.

What I have done so far:

$$\frac{(x^2-\sqrt{x^4+7x^2+1})(x^2+\sqrt{x^4+7x^2+1})}{ x^2+\sqrt{x^4+7x^2+1}}$$

results in

$$\frac{-7x^2-1}{x^2+\sqrt{x^4+7x^2+1}}$$

factor out the $x^4$ under the radical, then divide numerator and denominator by $x^2$ to get

$$\frac{-7-1/x^2}{1 + \sqrt{1+7/x^2+1/x^4}}$$

at this point the limit as x approaches infinity would be -7/2 or -3.5.

Martin Sleziak
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theialux
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1 Answers1

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Let $y = \sqrt{x^4+7x^2+1}$ and $L=\lim (x^2-y)$. Throughout this answer $\lim$ means $\lim_{x\rightarrow \infty}$.

I'll first work on simplifying $y$ as follows. We can take an $x^2$ out of the square root , so that $y= x^2 \sqrt{1+7/x^2 + 1/x^4}$. We can expand the right side out by using the binomial expansion $\sqrt{1+\epsilon} = 1 + \epsilon/2 + H(\epsilon)$, where $H$ stands for the higher order terms in the expansion. Thus, $y = x^2(1+ \frac{7}{2x^2} + \frac{1}{2x^4} + H(x))$. Note that $H(x)$ contains large negative powers of $x$, from $x^{-4}$ onwards.

Now we can find $L$. We know that

\begin{equation} L = \lim (x^2 - y) \\ =\lim x^2 - x^2(1+ \frac{7}{2x^2} + \frac{1}{2x^4} + H(x)) \\ = \lim x^2 - (x^2 + \frac{7}{2} + \frac{1}{2x^2} + H(x)x^2) \\ = \lim 0 - 7/2 - H(x)x^2 \end{equation}

where $H(x)x^2$ contains only negative powers of $x$. As such the limit evaluates to $L = -7/2$.

This is very similar to what you have done, except that I simplified the algebra and swept all unwanted terms into $H$. The answer matches yours.

Ganesh
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