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?

score 2 · Answer 1 · answered Apr 18 '18 at 17:26

2

Hint

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

answered Apr 18 '18 at 17:26

Arnaldo

21,342

$\forall x>1, (x-1)e^x>0$ – Alexandros Voliotis Apr 18 '18 at 17:28
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For $x=+\infty$ it is $+\infty$ – Alexandros Voliotis Apr 18 '18 at 17:29
@AlexanderVoliotis: that's right! So, your answer is? – Arnaldo Apr 18 '18 at 17:32
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Well, $+\infty$! – Alexandros Voliotis Apr 18 '18 at 17:56

score 2 · Answer 2 · answered Apr 18 '18 at 17:32

2

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

answered Apr 18 '18 at 17:32

Dr. Sonnhard Graubner

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Factorizing the dominant term is always a good choice when in the doubt. – zwim Apr 18 '18 at 18:54

score 1 · Answer 3 · answered Apr 18 '18 at 17:28

1

Hint:

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

answered Apr 18 '18 at 17:28

cansomeonehelpmeout

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score 1 · Accepted Answer · answered Apr 18 '18 at 17:29

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 $x\to\infty$.

answered Apr 18 '18 at 17:29

thesmallprint

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Przemysław Scherwentke · Answer 5 · 2018-04-18T17:33:09.520

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

edited Apr 18 '18 at 17:33

answered Apr 18 '18 at 17:26

Przemysław Scherwentke

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

5 Answers5