I am trying to compute the limit here, i am not sure how to do it if i just plug it in or do i do the conjugate and solve? For some reason i got infinity as my answer yet I was told theres an actual solution to it. Perhaps someone can help me out here?
$\lim\limits_{x\rightarrow \infty}\sqrt{16x^2-5x+10{}}-4x$
Answer
To paraphrase Tom Lehrer: Rationalize, rationalize, rationalize.
We use the fact that $(a-b)(a+b) = a^2-b^2$. So if we multiply
$$\sqrt{16x^2-5x+10} - 4x$$
by
$$\sqrt{16x^2-5x+10}+4x$$
that will eliminate the square root. Unfortunately, we are not allowed to just multiply by whatever we please (that changes the function), so we must be sure that what we do, in total, does not amount to anything. So in addition to multiplying by $\sqrt{16x^2-5x+10}+4x$, we also divide by it so that it all amounts to multiplying by $1$ (which doesn't do anything). Then we can do the limit the usual way, by noting that the highest degree of $x$ that shows up in both numerator and denominator is $x$. (I know it looks like an $x^2$ in the denominator, but it is inside a square root, so it amounts to just $x$):
$$\begin{align*}
\lim_{x\to\infty}\left(\sqrt{16x^2-5x+10}-4x\right) &= \lim_{x\to\infty}\frac{(\sqrt{16x^2-5x+10}-4x)(\sqrt{16x^2-5x+10}+4x)}{\sqrt{16x^2-5x+10}+4x}\\
&= \lim_{x\to\infty}\frac{16x^2-5x+10 - (4x)^2}{\sqrt{16x^2-5x+10}+4x}\\
&= \lim_{x\to\infty}\frac{-5x+10}{\sqrt{16x^2-5x+10}+4x}\\
&=\lim_{x\to\infty}\frac{\frac{1}{x}(-5x+10)}{\frac{1}{x}(\sqrt{16x^2-5x+10}+4x)}\\
&= \lim_{x\to\infty}\frac{-5 + \frac{10}{x}}{\frac{1}{x}\sqrt{16x^2-5x+10}+4}\\
&= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{\frac{16x^2-5x+10}{x^2}}+4}\\
&= \lim_{x\to\infty}\frac{-5+\frac{10}{x}}{\sqrt{16 - \frac{5}{x}+\frac{10}{x^2}} + 4}\\
&= \frac{-5+0}{\sqrt{16-0+0}+4}\\
&= \frac{-5}{4+4}\\
&= -\frac{5}{8}.
\end{align*}$$
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