First slide

Introduction to limits

Question

$If f (x) = lim_{n \to \infty} \sum_{r = 1}^{n} 3^{r - 1} \sin^{3} \frac{x}{3^{r}} then lim_{x \to 0} \frac{f (x)}{x^{3}} =$

Difficult

A

0
B

$\frac{1}{6}$
C

$\frac{1}{24}$
D

Does not exists

Solution

$\begin{array}{l} 3^{r - 1} \sin^{3} \frac{x}{3^{r}} = \frac{3^{r - 1}}{4} (3^{} \sin \frac{x}{3^{r}} - \sin \frac{x}{3^{r - 1}}) = \frac{1}{4} (3^{r} \sin \frac{x}{3^{r}} - 3^{r - 1} \sin \frac{x}{3^{r - 1}}) \\ (\begin{matrix} ∵ \sin 3 θ = 3 \sin θ - 4 \sin^{3} θ \\ \Rightarrow \sin^{3} θ = \frac{1}{4} [3 \sin θ - \sin 3 θ]) \end{matrix} \\ Now, f (x) = \underset{n \to \infty}{\frac{1}{4} lim} \sum_{r = 1}^{n} (3^{r} \sin \frac{x}{3^{r}} - 3^{r - 1} \sin \frac{x}{3^{r - 1}}) \\ = \frac{1}{4} lim_{n \to \infty} [(3^{} \sin \frac{x}{3^{}} - 3^{0} \sin \frac{x}{3^{0}}) + (3^{2} \sin \frac{x}{3^{2}} - 3^{1} \sin \frac{x}{3^{1}}) + (3^{3} \sin \frac{x}{3^{3}} - 3^{2} \sin \frac{x}{3^{2}}) + . . . . . + (3^{n} \sin \frac{x}{3^{n}} - 3^{n - 1} \sin \frac{x}{3^{n - 1}})] \\ = \frac{1}{4} lim_{n \to \infty} [3^{n} \sin \frac{x}{3^{n}} - \sin x] = \frac{1}{4} lim_{n \to \infty} [\frac{\sin \frac{x}{3^{n}}}{\frac{x}{3^{n}}} x - \sin x] = \frac{x - \sin x}{4} \end{array}$
$\begin{array}{l} ∴ lim_{x \to 0} \frac{f (x)}{x^{3}} = lim_{x \to 0} \frac{x - [x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots .]}{4 x^{3}} \\ = lim_{x \to 0} [\frac{x^{3}}{6 \times 4 \times x^{3}} - \frac{x^{2}}{5! \times 4} + \dots . .] \\ = \frac{1}{24} - 0 + 0 \dots + 0 \\ = \frac{1}{24} \end{array}$

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