# Limits by definition.

• Jan 9th 2010, 02:42 PM
Also sprach Zarathustra
Limits by definition.
Prove by definition (epsilon-delta) the next limits:

1. lim(x-->inf) {(2x^2-11x-11)/(4x^2+11x-1)} = 1/2

2. lim(x-->1) {(x^3-3x)/(x^2-2x+1)} = -inf

Thanks!
• Jan 9th 2010, 02:57 PM
Drexel28
Quote:

Originally Posted by Also sprach Zarathustra
Prove by definition (epsilon-delta) the next limits:

1. lim(x-->inf) {(2x^2-11x-11)/(4x^2+11x-1)} = 1/2

2. lim(x-->1) {(x^3-3x)/(x^2-2x+1)} = -inf

Thanks!

For the first one note that, $\frac{2x^2-11x-11}{4x^2+11-1x}=\frac{2-\frac{11}{x}-\frac{11}{x^2}}{4+\frac{11}{x}-\frac{1}{x^2}}$. Now, $\left|2-\frac{11}{x^2}-\frac{11}{x}-2\right|=\left|\frac{11}{x^2}+\frac{11}{x}\right|\ leqslant\frac{22}{x}$. So, let $\varepsilon>0$ be given. Choosing $x>\frac{22}{\varepsilon}$ ensures that $\left|2-\frac{11}{x^2}-\frac{11}{x}\right|\leqslant\frac{22}{x}<\frac{22} {\frac{22}{\varepsilon}}=\varepsilon$. We may conclude that $2-\frac{11}{x^2}-\frac{11}{x}\to2$. Using similar techniques we may conclude that $4+\frac{11}{x}-\frac{1}{x^2}\to4$. We may therefore conclude that $\lim_{x\to\infty}\frac{2x^2-11x-11}{4x^2+11x-x}=\lim_{x\to\infty}\frac{2-\frac{11}{x}-\frac{11}{x^2}}{4+\frac{11}{x}-\frac{1}{x^2}}=\frac{2}{4}=\frac{1}{2}$

For the second one, note that $\frac{x^3-x}{x^2-2x+1}=2+x+\frac{2}{x-1}$ which for a sufficiently small neighborhood around $1$ is strictly less than $3+\frac{2}{x-1}$. So, let $T<0$ be given, then choosing $1>x>\frac{T-1}{T-3}$ ensures that $3+\frac{2}{x-1}<3+\frac{2}{\frac{T-1}{T-3}}=T$. The conclusion follows.

P.S. Technically, the second should be $\lim_{x\to 1^{-}}$
• Jan 9th 2010, 03:06 PM
Also sprach Zarathustra
\left|2-\frac{11}{x^2}-\frac{11}{x}-2\right|=\left|\frac{11}{x^2}+\frac{11}{x}\right|\ leqslant\frac{22}{x} Wy is that? |2-....-2| ???
• Jan 9th 2010, 03:07 PM
Drexel28
Quote:

Originally Posted by Also sprach Zarathustra
\left|2-\frac{11}{x^2}-\frac{11}{x}-2\right|=\left|\frac{11}{x^2}+\frac{11}{x}\right|\ leqslant\frac{22}{x} Wy is that? |2-....-2| ???

What??????????????????????????????????????????
• Jan 9th 2010, 03:22 PM
Also sprach Zarathustra
Sorry!
• Jan 9th 2010, 03:23 PM
Drexel28
Quote:

Originally Posted by Also sprach Zarathustra
Sorry!

Haha. Don't be sorry! I knew you just didn't look close enough!