### Home > CALC > Chapter 12 > Lesson 12.2.4 > Problem12-100

12-100.

Use Taylor polynomials to quickly find the limits below.

1. $\lim\limits_ { x \rightarrow 0 } \frac { x - \operatorname { tan } ^ { - 1 } x } { x ^ { 3 } }$

$\tan^{-1}(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-...$

$\lim\limits_{x\to 0}\frac{x-\Big(x-\frac{x^3}{3}+\frac{x^5}{5}-...\Big)}{x^3}$

1. $\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { ln } ( x + 1 ) } { x ^ { 2 } }$

$\lim\limits_{x\to 0}\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-...}{x^2}$

Use l'Hôpital's Rule once.

1. $\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { sin } ( x ^ { 3 } ) - x ^ { 3 } } { x ^ { 9 } }$

$\lim\limits_{x\to 0}\frac{x^3-\frac{(x^3)^3}{3!}+\frac{(x^3)^5}{5!}-...-x^3}{x^9}$

Several iterations of l'Hôpital's Rule can be used. Do you notice a pattern?

1. $\lim\limits_ { x \rightarrow 1 } \frac { \operatorname { ln } x } { x - 1 }$

This limit is for $x → 1$, so write the polynomial centered at $x = 1$.

$\lim\limits_{x\to 1}\frac{(x-1)-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-...}{x-1}$

1. $\lim\limits_ { x \rightarrow 0 } \frac { \operatorname { tan } x - x } { x ^ { 3 } }$

$\lim\limits_{x\to 0}\frac{x+\frac{x^3}{3}+\frac{2x^5}{15}+...-x}{x^3}$

1. $\lim\limits_ { x \rightarrow 0 } \frac { 1 } { \operatorname { sin } x } - \frac { 1 } { x }$

$\lim\limits_{x\to 0}\Bigg(\frac{1}{x-\frac{x^3}{3!}+\frac{x^5}{5!}-...}-\frac{1}{x}\Bigg)$

Next, rewrite the argument as a single complex fraction.