Using continuity to evaluate a limit of a composite function

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

However, I tried to solve this problem using my Graphics Calculator instead of completing the square. I got the zeros of $x^2 - 2x - 4$ to be $x_1 = 3.236$ and $x_2 = -1.236$

Therefore $x_1 ≥ 3.236$ and $x_2 ≥ -1.236$

Since $x_1 > x_2$ then,

Therefore $(x | x ≥ -1.236 )$ which is $~(x | x ≥ 1 - \sqrt{5} )~$ (sorry curly brackets were not working), however, my domain restriction excludes some other values shown in the solution. Dose anybody please know a way using my method to get those values?

Many thanks!

 

[anuttarasammyak]

In the process of $lim_{x\rightarrow 4}$ we put x in the region $(4-\epsilon, 4+\epsilon)$ for any small positive number $\epsilon$, which is in your {x| 3.236<x} but not in {x|x<-1.236}.

For an example of another exercise of $lim_{x\rightarrow -2}$ we put x in the region $(-2-\epsilon, -2+\epsilon)$ for any small positive number $\epsilon$, which is not in your {x| 3.236<x} but is in {x|x<-1.236}.

 

[member 731016]

In the process of $lim_{x\rightarrow 4}$ we put x in the region $(4-\epsilon, 4+\epsilon)$ for any small positive number $\epsilon$, which is in your {x| 3.236<x} but not in {x|x<-1.236}.

For an example of another exercise of $lim_{x\rightarrow -2}$ we put x in the region $(-2-\epsilon, -2+\epsilon)$ for any small positive number $\epsilon$, which is not in your {x| 3.236<x} but is in {x|x<-1.236}.

Thank you for your reply @anuttarasammyak !

Sorry, I have not really done by epsilon-delta definition of a limit.

Many thanks!

 

[pasmith]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 324502
View attachment 324503
The solution is,
View attachment 324504
However, I tried to solve this problem using my Graphics Calculator instead of completing the square. I got the zeros of $x^2 - 2x - 4$ to be $x_1 = 3.236$ and $x_2 = -1.236$

Therefore $x_1 ≥ 3.236$ and $x_2 ≥ -1.236$

Since $x_1 > x_2$ then,

Therefore $(x | x ≥ -1.236 )$ which is $~(x | x ≥ 1 - \sqrt{5} )~$ (sorry curly brackets were not working), however, my domain restriction excludes some other values shown in the solution. Dose anybody please know a way using my method to get those values?

Many thanks!

For P(x) = ax^2 + bx + c:

If P has only a single real root, or no real roots:

• If a &gt; 0 then P(x) \geq 0 everywhere.
• If a &lt; 0 then P(x) \geq 0 only at the real root (if any).

If P has two distinct real roots r_1 &lt; r_2:

• If a &gt; 0 then P(x) \geq 0 for x \leq r_1 or x \geq r_2 ("outside the roots").
• If a &lt; 0 then P(x) \geq 0 for r_1 \leq x \leq r_2 ("between the roots").

All of these results follow from the facts that P(x) is positive for all sufficiently large |x| if a &gt; 0 and negative for all sufficiently large |x| if a &lt; 0 and that a polynomial only changes sign at a real root of odd multiplicity.

Alternatively, by completing the square you can reduce this to either |x - p| \leq q or |x - p| \geq q. The first has the interpretation that x is at most q units away from p, ie. p - q \leq x \leq p + q. The second has the interpretation that x is at least q units away from p, ie. x \leq p -q or x \geq p + q.

 

[WWGD]

This seems a case of using continuity implies sequential continuity. ( Though the converse is false).