1. ## Finding Delta

Use the precise definition to prove the following limit. Given epsilon, find a suitable delta to prove
lim 5x^2-3x-2 = 34
x->-3

So I got
5x^2-3x-36
|x+3| * |5x-12|

-1 < x+3 < 1
-4 < x < -2
-20 < 5x < -10
-32 < 5x-12 < -22
How do I find delta if my delta is negative?

Need help with these two as well.

Tried multiplying the conjugate with this one but ran into a dead end.
lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
x -> 0

lim (x^(2/3)-4) / (x^(1/3)-2)
x -> 8

Thanks for the help!

2. Originally Posted by beanus
Use the precise definition to prove the following limit. Given epsilon, find a suitable delta to prove
lim 5x^2-3x-2 = 34
x->-3

So I got
5x^2-3x-36
|x+3| * |5x-12|

-1 < x+3 < 1
-4 < x < -2
-20 < 5x < -10
-32 < 5x-12 < -22
How do I find delta if my delta is negative?

Need help with these two as well.

Tried multiplying the conjugate with this one but ran into a dead end.
lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
x -> 0

lim (x^(2/3)-4) / (x^(1/3)-2)
x -> 8

Thanks for the help!
Never mind you did. I couldn't tell at first.

Here are some worked out examples

http://www.math.ucdavis.edu/~kouba/C...l#SOLUTION%202

3. First of all, $\displaystyle \lim_{x \to -3}(5x^2 - 3x - 2) \neq 34$, it equals $\displaystyle 52$.

However, $\displaystyle \lim_{x \to 3}(5x^2 - 3x - 2) = 34$.

So, to prove $\displaystyle \lim_{x \to 3}(5x^2 - 3x - 2) = 34$, you need to show that $\displaystyle |x - 3| < \delta \implies |5x^2 - 3x - 2 - 34| < \epsilon$ for all $\displaystyle \epsilon > 0$.

So solve $\displaystyle |5x^2 - 3x - 2 - 34| < \epsilon$ for $\displaystyle |x - 3|$.

$\displaystyle |5x^2 - 3x - 36| < \epsilon$

$\displaystyle |x-3||5x+12| < \epsilon$.

$\displaystyle |x - 3| < \frac{\epsilon}{5x + 12}$.

Now, since taking a limit requires us to close in on the point $\displaystyle x = 3$, we can define $\displaystyle \delta$ to be a small value, e.g. $\displaystyle 1$, since we are going to make it smaller anyway...

So by defining $\displaystyle \delta = 1$, we have

$\displaystyle |x - 3| < 1$

$\displaystyle -1 < x-3 < 1$

$\displaystyle -5 < 5x - 15 < 5$

$\displaystyle 10 < 5x < 20$

$\displaystyle 22 < 5x + 12 < 32$

So we can say $\displaystyle |x - 3| < \frac{\epsilon}{32}$, provided $\displaystyle |x - 3| < 1$.

So define $\displaystyle \delta = \min\left\{1, \frac{\epsilon}{32}\right\}$, and the proof will follow.

4. Originally Posted by beanus
Need help with these two as well.

Tried multiplying the conjugate with this one but ran into a dead end.
lim (Sqrt(2+x^2) - Sqrt(2-x^2)) / x^2
x -> 0

lim (x^(2/3)-4) / (x^(1/3)-2)
x -> 8
For these are you expected to use the precise definition?

5. They just say "Find:"

6. $\displaystyle \lim_{x \to 0}\frac{\sqrt{2+x^2} - \sqrt{2-x^2}}{x^2} = \lim_{x \to 0}\frac{(\sqrt{2+x^2}-\sqrt{2-x^2})(\sqrt{2+x^2}+\sqrt{2-x^2})}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}$

$\displaystyle = \lim_{x \to 0}\frac{(2+x^2)-(2-x^2)}{x^2(\sqrt{2+x^2}+\sqrt{2-x^2})}$

$\displaystyle = \lim_{x \to 0}\frac{2x^2}{x^2(\sqrt{2+x^2} + \sqrt{2-x^2})}$.

I'm sure you can go from here...

7. Thanks! I get that one. Any thing on the last one? I multiplied by the conjugate and got

(x^4/3 - 16) / ((x^1/3 - 2) (x^2/3 + 4))

Note that $\displaystyle x^{\frac{2}{3}} - 4 = \left(x^{\frac{1}{3}}\right)^2 - 2^2 = \left(x^{\frac{1}{3}} - 2\right)\left(x^{\frac{1}{3}} + 2\right)$.