CPM Homework Help : CALC Problem 12-100

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12-100.

Use Taylor polynomials to quickly find the limits below.

$\lim\limits_ { x \rightarrow 0 } \frac { x - \operatorname { tan } ^ { - 1 } x } { x ^ { 3 } }$
Step 1(a):
$\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-...$
Step 2(a):
$\lim\limits_{x\to 0}\frac{x-\Big(x-\frac{x^3}{3}+\frac{x^5}{5}-...\Big)}{x^3}$

$\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { ln } ( x + 1 ) } { x ^ { 2 } }$
Step (b):
$\lim\limits_{x\to 0}\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-...}{x^2}$
More Help (b):
Use l'Hôpital's Rule once.

$\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { sin } ( x ^ { 3 } ) - x ^ { 3 } } { x ^ { 9 } }$
Step (c):
$\lim\limits_{x\to 0}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...-x^3}{x^9}$
More Help (c):
Several iterations of l'Hôpital's Rule can be used. Do you notice a pattern?

$\lim\limits_ { x \rightarrow 1 } \frac { \operatorname { ln } x } { x - 1 }$
Hint (d):
This limit is for $x → 1$ , so write the polynomial centered at $x = 1$ .
Step (d):
$\lim\limits_{x\to 1}\frac{(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-...}{x-1}$

$\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { tan } x - x } { x ^ { 3 } }$
Step (e):
$\lim\limits_{x\to 0}\frac{x+\frac{x^3}{3}+\frac{2x^5}{15}+...-x}{x^3}$

$\lim\limits_ { x \rightarrow 0 } \frac { 1 } { \operatorname { sin } x } - \frac { 1 } { x }$
Step (f):
$\lim\limits_{x\to 0}\Bigg(\frac{1}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}-\frac{1}{x}\Bigg)$
More Help (f):
Next, rewrite the argument as a single complex fraction.

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