Question

check my work

Answer 1

where is it?

Answer 2

Prove that
\[\lim_{x \rightarrow4} x^2-5=11\]
given
\[\epsilon=1\]
We have
to prove the statement
\[0<|x-x_{o}|<\delta \implies |f(x)-L|<\epsilon\]
\[0<|x-4|<\delta \implies |x^2-5-11|<1\]
\[0<|x-4|<\delta \implies |x^2-16|<1\]
Consider the inequality
\[|x^2-16|<1\]\[-1<x^2-16<1\]\[15<x^2<17\]\[\sqrt{15}<|x|<\sqrt{17}\]
In the near neighbourhood of
\[x_{o}=4\]
We have
\[|x|=x\]
thus
\[\sqrt{15}<x<\sqrt{17}\]
Now we have to fit the inequality
\[4-\delta<x<4+\delta\]
Into the interval (sqrt(15),sqrt(17))
For this we equate the left and right sides of both inequalities
\[4-\delta=\sqrt{15} \space \space , \space \space 4+\delta=\sqrt{17}\]\[\delta=4-\sqrt{15} \space \space , \space \space \delta=\sqrt{17}-4\]
We thus take delta to be the minimum of the values
\[\delta=\min(4-\sqrt{15},\sqrt{17}-4)=\sqrt{17}-4\]
Thus for \[\delta=\sqrt{17}-4\]
or any smaller positive value, the inequality
\[0<|x-4|<\delta\]
will automatically make
\[\sqrt{15}<x<\sqrt{17}\]
Hold good but this inequality is the same as
\[|x^2-16|<1\]
Thus we have found delta greater than 0 for some given epsilon greater than 0 such that
\[0<|x-4|<\delta \implies |x^2-16|<1\]
\[\implies \lim_{x \rightarrow 4}x^2-5=11\]

Answer 3

perfect :D

Answer 4

thanks