How to find $\lim _{x\to \infty }(3-4x)e^{-x}$ without L'Hôpitals?

Question

$$\lim _{x\to \infty }(3-4x)e^{-x} = \lim _{x\to \infty }-\infty * 0 $$

Since I'm not allowed to multiply infinity by zero, how am I supposed to solve this limit?

Answer 1

For $x > e$ say, $e^x > x^2$ and $\displaystyle {1 \over e^x} < {1 \over x^2}$. Hence for $x > e$,

$$0 < \left| {3-4x \over e^x }\right| < \left|{3 - 4x \over x^2}\right|$$

Now squeeze.

Answer 2

$$\frac{3-4x}{e^x}=\frac{3}{1+x+\frac{x^2}{2}+...+\frac{x^n}{n!}+..}-\frac{4}{\frac{1}{x}+1+\frac{x}{2}+...+\frac{x^{n-1}}{n!}+...}$$