Taylor Poly's and l'Hop

**Belzelga** · May 6th 2008, 12:41 AM

Could somebody help me with this problem?

Find the Taylor polynomials the 2nd Taylor polynomial $\begin{aligned} T_{f,0}^{(2)} (x)\end{aligned}$ around 0 for the following function

f(x) = (1)/(1+x)

and this nasty question...

Using l'Hoptal's rule...

lim (x->infin) [((cos) (a/x))^x^2] where a>0
lim (x->0) [((log(1+ax))/x) where a>0

**mr fantastic** · May 6th 2008, 03:02 AM

Originally Posted by Belzelga

Could somebody help me with this problem?

Find the Taylor polynomials the 2nd Taylor polynomial $\begin{aligned} T_{f,0}^{(2)} (x)\end{aligned}$ around 0 for the following function

f(x) = (1)/(1+x) Mr F says: This looks like a routine application of the formula. Where are you stuck?

and this nasty question...

Using l'Hoptal's rule...

lim (x->infin) [((cos) (a/x))^x^2] where a>0
lim (x->0) [((log(1+ax))/x) where a>0

$\lim_{x \rightarrow \infty} \left[ \cos \left( \frac{a}{x} \right) \right]^{x^2} = \lim_{t \rightarrow 0} \left[ \cos (at) \right]^{1/t^2} = \lim_{t \rightarrow 0} \, e^{\frac{\ln \cos (at)}{t^2}} = e^{\lim_{t \rightarrow 0}{\frac{\ln \cos (at)}{t^2}}}$ ......

---------------------------------------------------------------------------------------------

$\lim_{x \rightarrow 0} \frac{\ln(1+ax)}{x}$ is a routine application of l'Hopital's Rule if you're feeling lazy.

Otherwise you can substitute the Maclaurin series for $\ln (1 + ax) = ax - \frac{a^2x^2}{2} + ......$ :

$\lim_{x \rightarrow 0} \frac{ax - \frac{a^2x^2}{2} + .....}{x} = a$ .

**Belzelga** · May 6th 2008, 02:48 PM

Hi Fant

Just realised the first part was simple apparently so I got the following with few other examples:
===
Using l'hopital rule to compute limits
===
lim x->0 (1 - cos x)/ x^2

lim x->0 (sin x)/ 2x
lim x->0 (sin x)/ x = 1
lim x->0 (cos x)/ 2 = 1/2
===
===
lim x->0 (a^x - 1)/ k*x
lim x->0 [xa^(x-1) - 1]/ k = 1
===
===
And the ending of
lim x->infin (cos (a/x))^(x^2)

Is the limit 1/2?
===================

Now the Taylor Polynomial

f(x) = (1)/(1+x)
f'(x) = (-1)(1+x)^-2
f''(x) = (-2)(1+x)^-\3

f(0) = 1
f'(0) = -1
f''(0) = -2

Which will gave me (1/0!)-(1/1!)x-(2/2!)x^2 = 1-x-x^2
====

How did I do?

**TheEmptySet** · May 6th 2008, 02:58 PM

Originally Posted by Belzelga

Hi Fant

Now the Taylor Polynomial

f(x) = (1)/(1+x)
f'(x) = (-1)(1+x)^-2
f''(x) = (-2)(1+x)^-\3

f(0) = 1
f'(0) = -1
f''(0) = -2

Which will gave me (1/0!)-(1/1!)x-(2/2!)x^2 = 1-x-x^2
====

How did I do?

$f''(x)=(-2)(-1)(1+x)^{-3}$

$f''(0)=2$

**TheEmptySet** · May 6th 2008, 03:01 PM

Also note that

$\frac{1}{1+x}=\frac{1}{1-(-x)}$

This is the form of the geometric series with r=-x

so we get

$\frac{1}{1+x}=\sum_{n=0}^{\infty}(-x)^n=\sum_{n=0}^{\infty}(-1)^n(x)^n=1-x+x^2-x^3...$

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