# Thread: Prove that the limit exists using eps-delta definition

1. ## Prove that the limit exists using eps-delta definition

Question 1 )

limx->4 (7-3 x) = -5

I was able to solve it and found that d= -E/3 - But my problem is that how do I verify/show that my delta works? Please explain.

And there is another question where I am not sure how to approach it after simplifying it

Question 2)

lim x->2 (x^3)=8

|x-2||x^2+2x+4|<E |x-2|<d

2. Originally Posted by dennosan
Question 1 )

limx->4 (7-3 x) = -5

I was able to solve it and found that d= -E/3
Surely, you mean $\displaystyle \delta = \varepsilon/3$. Delta needs to be positive.

But my problem is that how do I verify/show that my delta works?
Well, how then did you find it?

In order to prove that $\displaystyle \lim_{x\to 4}(7-3x)=-5$, you need to show that for any $\displaystyle \varepsilon>0$ there exists a $\displaystyle \delta>0$ such that for all x with $\displaystyle |x-4|<\delta$ the following inequality holds:
$\displaystyle |(7-3x)-(-5)| < \varepsilon$
Now, simplify the left side to get
$\displaystyle |12-3x|< \varepsilon$
Pull out a positive factor of 3 like this
$\displaystyle 3\cdot |4-x|<\varepsilon$
and divide by 3 to get
$\displaystyle |4-x|<\varepsilon/3$
which is equivalent to
$\displaystyle |x-4|<\varepsilon/3$
Now all these inequalities are equivalent. So if the last inequality is satisfied, which is the case if you set $\displaystyle \delta := \varepsilon/3$ and require that $\displaystyle |x-4|<\delta$, then the first of these inequalities, the one that you have to show, also holds.

And there is another question where I am not sure how to approach it after simplifying it

Question 2)

lim x->2 (x^3)=8

|x-2||x^2+2x+4|<E |x-2|<d
$\displaystyle |x^3-8|<\varepsilon$
and try to figure out how requiring $\displaystyle |x-2|<\delta$, for a suitably chosen $\displaystyle \delta>0$, guarantees that this inequality holds.
$\displaystyle |x-2|\cdot |x^2+2x+4|<\varepsilon$
Well, $\displaystyle x^2+2x+4=(x+1)^2+3$ by completion of the square. Thus the quadratic polynomial and its absolute value are the same (because the polynomial is always positive) and it assumes its smallest value 3 at x=-1.
So lets therefore already limit x to satisfy |x-2|<1, i.e. x must be from the interval (1,3). For such values of x, the quadratic factor assumes its largest value at x=3, which gives us the required upper bound $\displaystyle M := 3^2+2*3+4=19$.
Finally, divide the above inequality by M on both sides to get your $\displaystyle \delta :=\min(1,\varepsilon/19)$.
(We must make sure that $\displaystyle \delta < 1$, because otherwies our choice of M as an upper bound for the quadratic factor on the left side of the inequality would not be valid.)