A quick way to find $\lim\limits_{x \to -\infty} \frac{8-8x}{14+x} + \frac{3x^2+7}{(8x-4)^2}$?

Question

I am not sure if there's a quick way to calculate $\lim\limits_{x \to -\infty} \frac{8-8x}{14+x} + \frac{3x^2+7}{(8x-4)^2}$.

I could combine the fractions, but there are three other limits included in the problem I'm doing, and so it seems like there might be a quicker way to do this.

Answer 1

It's simple: a theorem asserts that the limit at infinity of a rational function is the limit of the ratio of the leading terms of the numerator and the denominator.

In formula: $$\lim_{x\to-\infty}\frac{8-8x}{14+x} =\lim_{x\to-\infty}\frac{-8x}x=-8,\qquad \lim_{x\to-\infty}\frac{3x^2+7}{(8x-4)^2}=\lim_{x\to-\infty}\frac{3x^2}{(8x)^2}=\frac{3}{64}.$$

Answer 2

Note that $$\frac{8-8x}{14+x} = \frac{8/x-8}{14/x+1} \to -8$$

$$\frac{3x^2+7}{(8x-4)^2} = \frac{3+7/x^2}{(8-4/x)^2} \to \frac{3}{64}$$

when $x \to - \infty$.

Answer 3

$\dfrac{8-8x}{14+x}+\dfrac{3x^2+7}{(8x-4)^2}=\dfrac{\dfrac{8}{x}-8}{\dfrac{14}{x}+1}+\dfrac{3+\dfrac{7}{x^2}}{(8-\dfrac{4}{x})^2}\to-8+\dfrac{3}{8^2}$