Evaluate the following limit : limx→0[cos(ax)-cos(bx)cos(cx)-1] - Mathematics and Statistics

प्रश्न

Evaluate the following limit :

$lim_{x \to 0} [\frac{\cos (a x) - \cos (b x)}{\cos (c x) - 1}]$

बेरीज

उत्तर

$lim_{x \to 0} \frac{\cos (a x) - \cos (b x)}{\cos (c x) - 1}$

= $lim_{x \to 0} \frac{- \cos (a x) + \cos (b x)}{1 - \cos (c x)}$

= $lim_{x \to 0} \frac{1 - \cos a x - 1 + \cos b x}{1 - \cos c x}$

= $lim_{x \to 0} \frac{(1 - \cos a x) - (1 - \cos b x)}{1 - \cos c x}$

= $lim_{x \to 0} \frac{2 \sin^{2} (\frac{a x}{2}) - 2 \sin^{2} \frac{b x}{2}}{2 \sin^{2} \frac{c x}{2}}$

= $lim_{x \to 0} \frac{\frac{\sin^{2} (\frac{a x}{2}) - \sin^{2} (\frac{b x}{2})}{x^{2}}}{\frac{\sin^{2} (\frac{c x}{2})}{x^{2}}} ... [\begin{matrix} Divide numerator and denominator by x^{2} \\ ∵ x \to 0, ∴ x \neq 0 ∴ x^{2} \neq 0 \end{matrix}]$

= $\frac{lim_{x \to 0} [\frac{\sin^{2} (\frac{a x}{2})}{x^{2}} - \frac{\sin^{2} (\frac{b x}{2})}{x^{2}}]}{lim_{x \to 0} \frac{\sin^{2} (\frac{c x}{2})}{x^{2}}}$

= $\frac{lim_{x \to 0} {[\frac{\sin \frac{a x}{2}}{x}]}^{2} - lim_{x \to 0} {[\frac{\sin \frac{b x}{2}}{x}]}^{2}}{lim_{x \to 0} {[\frac{\sin \frac{c x}{2}}{x}]}^{2}}$

= $\frac{lim_{x \to 0} {[\frac{\sin \frac{a x}{2}}{\frac{a x}{2}}]}^{2} \cdot {(\frac{a}{2})}^{2} - lim_{x \to 0} {[\frac{\sin \frac{b x}{2}}{\frac{b x}{2}}]}^{2} \cdot {(\frac{b}{2})}^{2}}{lim_{x \to 0} {[\frac{\sin \frac{c x}{2}}{\frac{c x}{2}}]}^{2} \cdot {(\frac{c}{2})}^{2}}$

= $\frac{{(1)}^{2} \cdot \frac{a^{2}}{4} - {(1)}^{2} \cdot \frac{b^{2}}{4}}{{(1)}^{2} \cdot \frac{c^{2}}{4}}$

= $\frac{\frac{a^{2}}{4} - \frac{b^{2}}{4}}{\frac{c^{2}}{4}}$

= $\frac{a^{2} - b^{2}}{c^{2}}$

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Limits of Trigonometric Functions

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Q III. (1)Q II. (3)Q III. (2)

पाठ 7: Limits - Exercise 7.4 [पृष्ठ १४८]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 11 Standard Maharashtra State Board

पाठ 7 Limits
Exercise 7.4 | Q III. (1) | पृष्ठ १४८

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Evaluate the following limit : limx→0[cos(ax)-cos(bx)cos(cx)-1] - Mathematics and Statistics

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