MA101.15 A bit more on limits


\begin{aligned}   x&>0 \\ \frac{e^x}{x^n}&=\frac{1+x+\frac{x^2}{2}+...}{x^n} \\ &> \displaystyle\frac{\frac{x^m}{m!}}{x^n} \end{aligned}

for any possible integer m we care to choose.

Choose m_0>n then \displaystyle \frac{e^x}{x^n} > \frac{x^{m_0-n}}{m_0!}.

But \displaystyle \frac{x^{m_0-n}}{m_0!} \to \infty as x \to \infty.

So, \displaystyle \frac{e^x}{x^n}\to\infty as x\to\infty.

So it can be said that e^x increases more rapidly than any power of x.

Also (for n>0).

\begin{aligned} \displaystyle \lim_{x\to\infty} \frac{\ln{x}}{x^n} &= \lim_{y\to\infty}\frac{y}{e^{ny}} \\&= \lim_{z\to\infty} \frac{z}{ne^z} = 0 \;\mbox{from above.} \end{aligned}

Where x=e^y as x\to\infty, y\to\infty, and z=ny as y\to\infty, z\to\infty as n>0.

So it can be said that \ln{x} increases more slowly than any positive power of x.

Also (for n>0).

\begin{aligned} \displaystyle \lim_{x\to 0^+} x^n\cdot\ln{x} &= \lim_{y\to\infty}\frac{1}{y^n}\cdot\ln{\frac{1}{y}} \\&= \lim_{y\to\infty}\frac{-\ln{y}}{y^n} = 0 \;\mbox{from above.} \end{aligned}

Where x=\frac{1}{y}, y=\frac{1}{x}, x\to0^+, y\to\infty.


\begin{aligned} \displaystyle \lim_{x\to\infty}(e^x-x^4) &= \lim_{x\to\infty}(e^4(\frac{e^x}{x^4}-1)) \\&= \infty \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to\infty}\frac{e^x(-2x^2+3)}{3x^4+2x^2+6} &= \lim_{x\to\infty}\frac{e^x(-2+\frac{3}{x^2})}{3x^2+2+\frac{6}{x^2}} \\&= -\infty \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to\infty}\frac{2e^{2x}+x^2+2}{3e^{3x}+7} &= \lim_{x\to\infty}\frac{2e^{-x}+\frac{x^2}{e^{3x}}+\frac{2}{e^{3x}}}{3+\frac{7}{e^{3x}}} \\&= 0 \end{aligned}

\begin{aligned} \displaystyle \lim_{x\to\infty} \frac{(x-1)\ln{x^2}}{x^2} &= \lim_{y\to\infty}\frac{(y^\frac{1}{2}-1)\ln{y}}{y} \\&= \lim_{y\to\infty}(\frac{\ln{y}}{y^\frac{1}{2}}-\frac{\ln{y}}{y}) \\&= 0 \end{aligned}

x^2=y, x\to\infty, y\to\infty.

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