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## Homework Statement

Use Taylor's remainder formula to show that the Taylor series for f(x) is about the point indicated converges to f(x) for all x.

[tex]f(x) = e^{5x}[/tex] about [tex]x=0[/tex]

## Homework Equations

## The Attempt at a Solution

Since,

[tex]f^{n}(x) = 5^{n}e^{5x}[/tex],

Taylor's remainder formula for [tex]e^{5x}[/tex] and c = 0 gives

[tex]e^{5x} = 1 + 5x + \frac{5^{2}}{2!}x^{2} + \frac{5^{3}}{3!}x^{3} + ... + \frac{5^{n}}{n!}x^{n} + R_{n}[/tex]

Where [tex]R_{n} = \frac{5^{n+1}e^{5z_{n}}}{(n+1)!}x^{n+1}[/tex]

There are two cases we must consider:

**Case 1: If x < 0 then,**

******NOTE: The following line is the line I am confused about,******

[tex]x < z_{n} < 0,[/tex] and [tex]|R_{n}| < 5^{n+1}\frac{|x|^{n+1}}{(n+1)!}[/tex]

How do they go about making this conclusion? Do they work from the this in the following manner?

[tex]x < z_{n} < 0[/tex]

[tex]5x < 5z_{n} < 0[/tex]

[tex]e^{5x} < e^{5z_{n}} < e^{0}[/tex]

[tex]e^{5x}\frac{|x|^{n+1}}{(n+1)!} < e^{5z_{n}}\frac{|x|^{n+1}}{(n+1)!} < e^{0}\frac{|x|^{n+1}}{(n+1)!}[/tex]

And then take the limit as n goes to infinity on the outside and use squeeze theorem to prove the rest?

Where are they getting the second part of this line:

Also, they are using squeeze theorem around [tex]lim_{n \rightarrow \infty} |R_{n}|[/tex], how come they are only showing one function around [tex]|R_{n}|[/tex]. (i.e. [tex]5^{n+1}\frac{|x|^{n+1}}{(n+1)!}[/tex]) Where is the 0 term?[tex]x < z_{n} < 0,[/tex] and [tex]|R_{n}| < 5^{n+1}\frac{|x|^{n+1}}{(n+1)!}[/tex]

Once I understand this I will continue and finish off the problem, this is the only line that confuses me.

Can someone please clarify?

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