Evaluating $\lim_{x\to-\infty} \frac{ 8x + 6}{\sqrt{x^2+8x} - x}$

Question

Given the function $$f(x) = \frac{ 8x + 6}{\sqrt{x^2+8x} - x}$$ what is $\lim_{ x \rightarrow -\infty } f(x)$?

Now, the correct solution to the problem suggests the following:

Let $-x = t,$ then $$\begin{align} \lim_{ x \rightarrow -\infty } f(x) &= \lim_{ x \rightarrow -\infty } \frac{ 8x + 6}{\sqrt{x^2+8x} - x} \\ &= \lim_{ t \rightarrow \infty } \frac{ -8t + 6}{\sqrt{t^2-8t} + t} \\ &= \lim_{ t \rightarrow \infty } \frac{ -8 + \frac{6}{t}}{\sqrt{1-\frac{8}{t}} + 1}. \end{align}$$

Thus, $\lim_{ x \rightarrow -\infty }f(x) = \dfrac{-8}{1+1} = -4$. $\square$

I do not understand, however, why we have to substitute $-x$ for $t$ instead of rationalizing the expression and solving as it is, and why actually doing so produces a different answer: $$\begin{align} \lim_{ x \rightarrow -\infty } f(x) &= \lim_{ x \rightarrow -\infty } \frac{(8x + 6)(\sqrt{x^2 + 8x} + x)}{x^2+8x-x^2} \\ &= \lim_{ x \rightarrow -\infty } \frac{(8x+6)(\sqrt{1+\frac{8}{x}}+1)}{8} \\ &= \lim_{ x \rightarrow -\infty } \frac{16x+12}{8} \\ &=-\infty \end{align}$$

Would anyone be so kind as to point out my mistake?

Answer 1

$\sqrt{x^2+8x}+x=|x|\sqrt{1+{8\over x}}+x$, if $x<0$ we obtain $\sqrt{x^2+8x}+x=-x\sqrt{1+{8\over x}}+x=-x(\sqrt{1+{8\over x}}-1)$

Answer 2

Steps:

$$\lim _{x\to \:-\infty \:}\left(\frac{8x+6}{\sqrt{x^2+8x}-x}\right)=\lim _{x\to \:-\infty \:}\frac{8+\frac{6}{x}}{-\sqrt{1+\frac{8}{x}}-1}$$ dividing by highest denominator power. After:

$$\lim _{x\to \:-\infty \:}\frac{8+\frac{6}{x}}{-\sqrt{1+\frac{8}{x}}-1}=\frac{\lim _{x\to \:-\infty \:}\left(8+\frac{6}{x}\right)}{\lim _{x\to \:-\infty \:}\left(-\sqrt{1+\frac{8}{x}}-1\right)}=-\frac82=-4$$

Keep in mind that $x\rightarrow -\infty$.

Answer 3

In your method, $x$ is large negative, $\dfrac{1}{x}\sqrt{x^{2}+8x}$ is not $\sqrt{1+\dfrac{8}{x}}$.

So you must get something like $-\dfrac{1}{-x}\sqrt{x^{2}+8x}=-\sqrt{1+\dfrac{8}{x}}$, but this does not really help to compute the limit, because you will end up with $-\sqrt{1+\dfrac{8}{x}}+1$, and when $x\rightarrow-\infty$, this goes to zero, but $8x+6$ goes to $-\infty$, so the result is of indeterminate form $-\infty\cdot 0$, which we fail to conclude anything from this.