How this kind of limit with variables in solved ?

[Martin117]

Homework Statement

[/b]

This is mentioned in my maths book, I m not able to understand it what they did.

Homework Equations

The Attempt at a Solution

 

[Martin117]

https://imgur.com/SxsGKsR here is the link to question

 

[Ray Vickson]

Homework Statement

[/B]
This is mentioned in my maths book, I m not able to understand it what they did.

Homework Equations

The Attempt at a Solution

They just factored out $(x-a)^p$ from both the numerator and the denominator (possibly with a different value of $p$ for the numerator and denominator.) Let $p_1 = p_{\text{numerator}}$ and $p_2 = p_{\text{denominator}}$.
(1) If $p_1 = p_2$ then all factors $(x-a)$ cancel out completely, and you are left with a fraction of the form $f_1(x)/g_1(x)$, where neither $f_1$ nor $g_1$ vanishes at $x = a$---so you get a finite, nonzero limit.
(2) If $p_1 > p_2$ your fraction simplifies to $(x-a)^{p_1 - p_2} \times f_1(x)/g_1(x)$, where neither $f_1$ nor $f_2$ vanishes at $x = a$. Therefore, the limit is $0$, because $(x-a)^{p_1 - p_2 } \to 0.$
(3) If $p_1 < p_2$ the fraction simplifies to $f_1(x)/f_2(x) \times 1/(x-a)^{p_2 - p_1}$. Since $1/(x-a)^{p_2 - p_1} \to \pm \infty$, the limit does not exist.

 

[Martin117]

They just factored out $(x-a)^p$ from both the numerator and the denominator (possibly with a different value of $p$ for the numerator and denominator.) Let $p_1 = p_{\text{numerator}}$ and $p_2 = p_{\text{denominator}}$.
(1) If $p_1 = p_2$ then all factors $(x-a)$ cancel out completely, and you are left with a fraction of the form $f_1(x)/g_1(x)$, where neither $f_1$ nor $g_1$ vanishes at $x = a$---so you get a finite, nonzero limit.
(2) If $p_1 > p_2$ your fraction simplifies to $(x-a)^{p_1 - p_2} \times f_1(x)/g_1(x)$, where neither $f_1$ nor $f_2$ vanishes at $x = a$. Therefore, the limit is $0$, because $(x-a)^{p_1 - p_2 } \to 0.$
(3) If $p_1 < p_2$ the fraction simplifies to $f_1(x)/f_2(x) \times 1/(x-a)^{p_2 - p_1}$. Since $1/(x-a)^{p_2 - p_1} \to \pm \infty$, the limit does not exist.

How they introduced new term (x-a)^k g1(x) and (x-a)^l h1(x) , from just g(a) and h(a) .

 

[Borek]

Just polynomial properties. Every polynomial can be expressed as $P(x) = c(x-a_1)(x-a_2)(x-a_3)...$ where a_i are its roots, and if so $P(x) = (x-a_1)Q(x)$ where $Q(x) = c(x-a_2)(x-a_3)...$. Powers are to allow for roots multiplicity.

 

[Martin117]

my confusion is how they replaced Lim_{x tends to a} g(x)=g(a) to g(x)=(x-a)^kg₁(x) and similarly with h(x) if this is a property of polynomial can you explain in detail because i m untouched to this .

 

[Borek]

Have you read what I wrote? Assume g(x) is P(x) and g₁(s) is Q(X). Do you see it now?

 

[Martin117]

Have you read what I wrote? Assume g(x) is P(x) and g₁(s) is Q(X). Do you see it now?

yeah now i m very near just last doubt please what is "c" in p(x) is this a constant term and if lim x tends to a g(x)=g(a) , g(x)=(x-a)^k g₁(x)
then g(a) =(x-a)^k g₁(x) what does it imply .

 

[Borek]

c is just a constant.