altri limiti notevoli dimostrati in R

Teorema

$a>1,k\in \mathbb{R}$

$\underset{x\to +\infty }{\lim }\frac{a^x}{x^k}=\underset{x\to +\infty }{\lim }{\left[\frac{{\left(a^{\frac{1}{k}}\right)}^x}{x}\right]}^k=+\infty$

poiché $\underset{x\to +\infty }{\lim }\frac{{\left(a^{\frac{1}{k}}\right)}^x}{x}=+\infty$

$\square$

---

$a>1,k\in \mathbb{R}$

$\underset{x\to -\infty }{\lim }{x}^k{a}^x$

$y=-x$

$\underset{x\to +\infty }{\lim }(-y{)}^k{a}^{-y}=\underset{x\to +\infty }{\lim }\frac{\stackrel{limitato}{\overbrace{(-1{)}^k}}{y}^k}{a^y}=0$

$\square$

---

$a>1$

$\underset{x\to +\infty }{\lim }x{\log }_{a}x$

$y={\log }_{a}x$
$x=a^y$

$\underset{x\to -\infty }{\lim }{a}^y\cdot t=0$
come dimostrato prima, prendendo $K=1$

$\square$

---

$\underset{x\to 0^{+}}{\lim }\sqrt{x}{\log }_{a}x$

$y={\log }_{a}x$
$x=a^y$

$\underset{x\to -\infty }{\lim }(a^{\frac{1}{2}}{)}^y\cdot y=0$

$\square$

---

$\underset{x\to +\infty }{\lim }\frac{{\log }_{a}x}{x}$

$y={\log }_{a}x$
$x=a^y$

$\underset{y\to +\infty }{\lim }\frac{y}{a^y}$

$z=-y$

$\underset{z\to +\infty }{\lim }-z{a}^z=0$

$\square$

---

$\underset{n}{\lim }(1+\frac{1}{n}{)}^n=e$

$\underset{x\to +\infty }{\lim }(1+\frac{1}{x}{)}^x\stackrel{?}{=}e$

$n=[x]$

$[x]\le x<[x]+1$
$\frac{1}{[x]+1}<\frac{1}{x}\le \frac{1}{[x]}$
$1+\frac{1}{[x]+1}<1+\frac{1}{x}\le 1+\frac{1}{[x]}$
${\left(1+\frac{1}{[x]+1}\right)}^{[x]}<{\left(1+\frac{1}{x}\right)}^x<{\left(1+\frac{1}{[x]}\right)}^{[x]+1}$
${\left(1+\frac{1}{n+1}\right)}^n<{\left(1+\frac{1}{x}\right)}^x<{\left(1+\frac{1}{n}\right)}^{n+1}$

$\frac{{\left(1+\frac{1}{n+1}\right)}^n}{1+\frac{1}{n+1}}<{\left(1+\frac{1}{x}\right)}^x<\left(1+\frac{1}{n}\right){\left(1+\frac{1}{n}\right)}^n$

$\frac{\stackrel{e}{\overbrace{{\left(1+\frac{1}{n+1}\right)}^n}}}{\underset{1}{\underbrace{1+\frac{1}{n+1}}}}<{\left(1+\frac{1}{x}\right)}^x<\underset{1}{\underbrace{\left(1+\frac{1}{n}\right)}}\underset{e}{\underbrace{{\left(1+\frac{1}{n}\right)}^n}}$

$\square$

---

$\underset{x\to -\infty }{\lim }{\left(1+\frac{1}{x}\right)}^x$

$y=-x$

$\underset{y\to +\infty }{\lim }{\left(1+\frac{1}{-y}\right)}^{-y}=\underset{y\to +\infty }{\lim }\frac{1}{{\left(1+\frac{1}{-y}\right)}^y}=\underset{y\to +\infty }{\lim }{\left(\frac{1}{1+\frac{1}{-y}}\right)}^y=\underset{y\to +\infty }{\lim }{\left(\frac{1}{\frac{-y+1}{-y}}\right)}^y=\underset{y\to +\infty }{\lim }{\left(\frac{y}{y-1}\right)}^y=\underset{y\to +\infty }{\lim }{\left(\frac{y+1-1}{y-1}\right)}^y=\underset{y\to +\infty }{\lim }{\left(1+\frac{1}{y-1}\right)}^y=\underset{y\to +\infty }{\lim }{\left(1+\frac{1}{y-1}\right)}^{y-1}\left(1+\frac{1}{y}\right)=e\cdot 1=e$

$\square$

---

$\underset{x\to 0}{\lim }\frac{\log (1+x)}{x}=\underset{x\to 0}{\lim }\log ((1+x{)}^{\frac{1}{x}})=\log e=1$

$\square$

---

$\underset{x\to 0}{\lim }\frac{e^x-1}{x}$

$y=e^x-1$
$x=\log (1+y)$

$\underset{y\to 0}{\lim }\frac{y}{\log (1+y)}=1$

$\square$

---

$\underset{x\to 0}{\lim }\frac{a^x-1}{x}=\underset{x\to 0}{\lim }\frac{e^{x\log a}-1}{x}\cdot \frac{\log a}{\log a}=1\cdot \log a=\log a$

$\square$

---

$\underset{x\to 0^{+}}{\lim }{x}^x=\underset{x\to 0^{+}}{\lim }{e}^{x\log x}=e^0=1$

$\square$

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