Baffled with calculating $\lim\limits_{x\to +\infty}{(2xe^x-2e^x+1)}$

Question

Given $$f(x)=2(x-1)e^x+1$$

Find: $\lim\limits_{x\to +\infty}{f(x)}$

Personal work:

$$\lim\limits_{x\to +\infty}{f(x)}=\lim\limits_{x\to +\infty}{(2xe^x-2e^x+1)}=\lim\limits_{x\to +\infty}({e^x \over 1/2x}-2e^x+1)$$... Gets to nowhere. Also, subtituing yields no results. Any suggestions?

Answer 1

$$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}e^x\Big(2x-2+\frac{1}{e^x}\Big)=\lim_{x\to\infty}e^x\cdot\lim_{x\to\infty}\Big(2x-2+\frac{1}{e^x}\Big)=\infty,$$ since the limit of $e^x$ is $\infty$ as $x\to\infty$.

Answer 2

Hint

What happens with $(x-1)e^x$ when $x$ is too big?

Answer 3

HINT: $$ \lim_{x\to\infty}2(x-1)e^x=\infty. $$ (Why?)

Answer 4

Hint:

What is: $$\lim_{x\rightarrow\infty}x=\dots\tag{1}$$ $$\lim_{x\rightarrow\infty}e^x=\dots\tag{2}$$

Answer 5

Hint: Write the term in the form $$xe^x\left(2-\frac{2}{x}+\frac{1}{xe^x}\right)$$