# Using continuity to evaluate a limit of a composite function

• ChiralSuperfields
In summary: Many thanks!In summary, the solution to this problem is that if a polynomial has two real roots, then it is positive or negative depending on whether a>0 or a<0.
ChiralSuperfields
Homework Statement
Relevant Equations
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!

Last edited:
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
anuttarasammyak said:
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}.

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

Many thanks!

ChiralSuperfields said:

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 $x \geq p + q$.

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

ChiralSuperfields

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