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?
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}