Thread: calculus help on Limits and epsilon delta

1. calculus help on Limits and epsilon delta

Hi, I do need your help with the question attached above.

I do have another question:

Prove by the eplison delta method

Lim of X^4 when x approaches -2 equal to 16

2. Originally Posted by guess
Hi, I do need your help with the question attached above.

I do have another question:

Prove by the eplison delta method

Lim of X^4 when x approaches -2 equal to 16
I'll do the $\displaystyle \varepsilon-\delta$ problem.

$\displaystyle \lim_{x\to-2}x^4=16$

The objective is to find $\displaystyle \delta$ such that $\displaystyle 0<\left|x^4-16\right|<\delta$ whenever $\displaystyle \left|x+2\right|<\varepsilon$.

Note that $\displaystyle \left|x^4-16\right|=\left|(x^2+4)(x^2-4)\right|=\left|(x^2+4)(x-2)(x+2)\right|<\delta$.

Let us supposed $\displaystyle \left|x+2\right|<1$. We find the upper bounds on $\displaystyle \left|x-2\right|$ and $\displaystyle \left|x^2+4\right|$

$\displaystyle \left|x-2\right|=\left|x+2-4\right|\leq\left|x+2\right|+\left|-4\right|=1+4=5$

$\displaystyle \left|x^2+4\right|=\left|x^2-4+8\right|\leq\left|x-2\right|\left|x+2\right|+\left|8\right|=(5)(1)+8=1 3$.

Thus, $\displaystyle \left|x^4-16\right|<13\left|x+2\right|<\varepsilon\implies \left|x+2\right|<\frac{\varepsilon}{13}$.

If we take $\displaystyle \delta=\frac{\varepsilon}{13}$, the proof is complete.

Does this make sense?

3. hi,

correct me if i'm wrong:

|X^4-16|= |(X+2)(X-2)(X^2+4)|

|X-2|=|X+2-4|<|X+2|+|-4|

we let delta less then 1

|X+2|< 1

|X-2| < (3)+(4) = 7

|X^2-4+8|<|(X+2)(X-2)+8|<|X+2||X-2|+|8| = (1)(5)+8 =13

from here,

|X^4-16|= |(X+2)(X-2)(X^2+4)|< delta(7)(13) = 91 delta

91 delta = epsilon
delta = epsilon/91.

We chose the min delta {1, epsilon/91}

4. Originally Posted by Chris L T521

If we take $\displaystyle \delta=\min\left\{1,\frac{\varepsilon}{13}\right\}$, the proof is complete