I tried:
$\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x}) = \frac{1}{x^2+1} \cdot e^xx = 0 \cdot \infty = 0$
But this is wrong. What did I do wrong? How do I solve this?
Find $\lim_{x \rightarrow \infty}(\frac{x}{x^2+1}\cdot e^{x})$
3
$\begingroup$
calculus
limits
limits-without-lhopital
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2Do you know that $\lim_{x \to +\infty} e^x/x = +\infty$? – 2017-02-22
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0Is the limit not $0$? – 2017-02-22
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1$xe^x$ dominates any polynomial. The limit is $+\infty$. – 2017-02-22
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1@Mark Beware: Zero times infinity is certainly not just zero. It can be anything at this point (indeterminate) – 2017-02-22
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0@imranfat You are absolutely right! lol My mistake! – 2017-02-22
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0What you did wrong is that you tried to apply the limit laws in the case $0 \times \infty$. The limit laws aren't applicable in that case. $0 \times \infty$ is called an "indeterminate form." – 2017-02-22
5 Answers
1
Unlike the well known standard limit $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$$ the following property of exponential function is not that popular/advertized: $$\lim_{x \to \infty}\frac{x^{n}}{e^{x}} = 0, n \in \mathbb{R}$$ It is this particular result which we need here. We have $$\lim_{x \to \infty}\frac{x}{x^{2} + 1}\cdot e^{x} = \lim_{x \to \infty}\frac{x^{2}}{x^{2} + 1}\cdot\frac{e^{x}}{x} = \infty$$
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0Because if $\lim_{x \to \infty}\frac{x^{n}}{e^{x}} = 0, n \in \mathbb{R}$, then it can be infered that $e^x > x^n$ and so $\frac{e^x}{x^n} \rightarrow \infty$. Thanks – 2017-02-23
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You're incorrect, as $0 \times \infty \neq 0$. It is indeterminate.
Note that as $e^t \ge t+1$ for all $t$, we have that $e^{\frac{x}{2}} \ge \frac{x}{2}+1$. Squaring both sides, we have that $$e^{x} \ge \frac{x^2}{4}+x+1>\frac{x^2}{4}+\frac{1}{4} $$ So, note that $$\frac{x}{x^2+1}e^x > \frac{x}{x^2+1} \times \left( \frac{x^2}{4}+\frac{1}{4}\right)=\frac{x}{4}$$ So $$\lim_{x \rightarrow \infty}\left(\frac{x}{x^2+1}\cdot e^{x}\right) \ge \lim_{x \to \infty}\frac{x}{4}$$ However, you can see that the limit on the right tends to infinity. So your limit will also tend to infinity.
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HINT:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\ge \frac{x^2}{2!}$$for $x\ge 0$
2
The biggest mistake in your working is assuming that $0 \cdot \infty = 0$. It is in fact an indeterminate form, so the answer isn't so simple.
One way of computing this would be $$\lim_{x\to\infty} \frac{x e^x}{x^2+1} = \lim_{x\to\infty} \frac{e^x}{x+\frac{1}{x}}$$ $$\geq \lim_{x\to\infty} \frac{e^x}{x+1}$$ $$=\infty$$
Since $\frac{1}{x}\leq1$, and the exponential function dominates any polynomial.
2
A slight different approach to S.C.B is to note that $e^x>1+x+\frac{x^2}{2}$ and so (worded informally) replacing the e-power in your limit and retaking the limit yields a cubic numerator and a quadratic denominator. Now $x$ goes to infinity, so...