Using continuity to evaluate a limit of a composite function

[ChiralSuperfields]

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}.

 

[ChiralSuperfields]

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 $a > 0$ then $P(x) \geq 0$ everywhere.
• If $a < 0$ then $P(x) \geq 0$ only at the real root (if any).

If $P$ has two distinct real roots $r_1 < r_2$:

• If $a > 0$ then $P(x) \geq 0$ for $x \leq r_1$ or $x \geq r_2$ ("outside the roots").
• If $a < 0$ 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 $a > 0$ and negative for all sufficiently large $|x|$ if $a < 0$ 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).