Small confusion about Taylor's remainder

[jegues]

Homework Statement

Please assume that what I have for the remainder is correct, and we are on the domain 2 < x < 4 around 2.

Homework Equations

The Attempt at a Solution

0 \leq |R_{n} (2,x)| = \frac{1}{n+1} |\frac{x-2}{z_{n}}|^{n+1}

Since,

2 &lt; x &lt; 4 then, 0 &lt; x-2 &lt; 2 and 0 &lt; \frac{x-2}{z_{n}} &lt; \frac{2}{z_{n}}

But, 2 &lt; z_{n} &lt; 4

So we can see that, \frac{2}{z_{n}} &lt; 1

Therefore, \frac{x-2}{z_{n}} &lt; 1.


Up to here makes perfect sense. Then he writes the following,

0 \leq |R_{n}(2,x)| &lt; \frac{1}{n+1} \cdot (1)^{n+1}

I'm confused about the, (1)^{n+1} how does he get that term? We showed that,

\frac{x-2}{z_{n}} &lt; 1, and if we take that and raise it to any power, say n+1, we will get 0. How does he get that 1 there?

Everything else makes perfect sense, it's just that one part.

Can someone please explain?

 

[Office_Shredder]

\frac{x-2}{z_{n}} &lt; 1, and if we take that and raise it to any power, say n+1, we will get 0.

You can't raise anything to any power and get zero. When you say we take "that", what are you referring to? 1ⁿ⁺¹=1 and \frac{x-2}{z_n}^{n+1} is going to be some small positive number

if a<b then aⁿ<bⁿ for any value of n (you can see this by noticing that the derivative of xⁿ is positive so is an increasing function)

 

[jegues]

You can't raise anything to any power and get zero. When you say we take "that", what are you referring to? 1ⁿ⁺¹=1 and \frac{x-2}{z_n}^{n+1} is going to be some small positive number

if a<b then aⁿ<bⁿ for any value of n (you can see this by noticing that the derivative of xⁿ is positive so is an increasing function)

The part in bold is what I was missing,

Since,

\frac{x-2}{z_{n}} &lt; 1 then,

\left( \frac{x-2}{z_{n}} \right)^{n+1} &lt; 1^{n+1}

So we can see that,

0 \leq |R_{n}(2,x)| &lt; \frac{1}{n+1} \cdot (1)^{n+1}

Applying squeeze theorem,

lim_{n \rightarrow \infty} 0 = lim_{n \rightarrow \infty} \frac{1}{n+1} = 0

Therefore,

lim_{n \rightarrow \infty} |R_{n}(2,x)| = 0

So it follows that,

lim_{n \rightarrow \infty} R_{n}(2,x) = 0 for 2 &lt; x &lt; 4