Infinity - Infinity indetermination in limits

Question

How can I solve this?

$$\lim_{x \to \infty} 2x - e ^{x - 1}$$

Answer 1

Note that

$$\lim_{x \to \infty} 2x - e ^{x - 1}=\lim_{x \to \infty} 2x\left(1 - \frac{e ^{x - 1}}{2x}\right)=(+\infty\cdot -\infty)=-\infty$$

indeed eventually $e^{x-1}\ge x^2$ and thus

$$\frac{e ^{x - 1}}{2x}\ge\frac{x^2}{2x}=\frac x 2 \to \infty$$

Answer 2

Let $[x-1]$ denote the largest integer not exceeding $x-1.$ We have $x-1\geq[x-1]>x-2.$ We have $e>2.$ By the Binomial Theorem we have $$x\geq 3\implies e^{x-1}>2^{x-1}\geq 2^{[x-1]}=(1+1)^{[x-1]}=$$ $$=\binom {[x-1]}{0}+\binom {[x-1]}{1}+\binom {[x-1]}{2}+....\;>$$ $$>\binom {[x-1]}{2}=$$ $$=\frac {[x-1]([x-1]-1)}{2}>$$ $$> \frac {(x-2)(x-3)}{2}.$$ Therefore $x\geq 3\implies 2x-e^{x-1}<2x- \frac {(x-2)(x-3)}{2}.$