# evaluating limits

• Jul 17th 2008, 02:21 PM
winterwyrm
evaluating limits
Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1
and lim(n->infinity) (ln(n))^(1/n) = 1

Thanks so much.
• Jul 17th 2008, 02:31 PM
Mathstud28
Quote:

Originally Posted by winterwyrm
Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1
and lim(n->infinity) (ln(n))^(1/n) = 1

Thanks so much.

I willl do the second one and you do the first. Ok?

We know that

$\forall{x}\geq{e}\quad{1}\leq\ln(x)$

Also we know that

$\forall{x}\in\mathbb{R}^+\quad{\ln(x)\leq{x}}$

$\Rightarrow\forall{x>e}\quad{1\leq\ln(x)\leq{x}}$

Now remaining inequality we see that

$\forall{x}>e\quad1\leq\ln(x)^{\frac{1}{x}}\leq{x^{ \frac{1}{x}}}$

Now we know that if we define $D=\left\{x:x\to\infty\right\}$

That and $R=\left\{x:x>e\right\}$

We can see that $D\subset{R}$

This implies that

$\lim_{x\to\infty}1\leq\lim_{x\to\infty}\ln(x)^{\fr ac{1}{x}}\leq\lim_{x\to\infty}x^{\frac{1}{x}}$

Now obviously $\lim_{x\to\infty}1=1$

and less obviously we can see that

$\lim_{x\to\infty}x^{\frac{1}{x}}\overbrace{=}^{\te xt{R/R relationsip}}=\lim_{x\to\infty}\frac{x+1}{x}=1$

$\Rightarrow\quad1\leq\lim_{x\to\infty}\ln(x)^{\fra c{1}{x}}\leq{1}$

$\Rightarrow\lim_{x\to\infty}\ln(x)^{\frac{1}{x}}=1 ~\text{By the Squeeze Thereom}\quad\blacksquare$
• Jul 17th 2008, 02:39 PM
ThePerfectHacker
Quote:

Originally Posted by winterwyrm
Use the squeeze theorem to show that lim(n->infinity) (5+2n+7(n^2))^(1/n) = 1

$1\leq 5+2n+7n^2 \leq 5n^2+2n^2+7n^2 = 14n^2 \implies 1 \leq (5+2n+7n^2)^{1/n} \leq 14^{1/n} \cdot (n^{1/n})^2$.