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.
Best Answer
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.