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Lim X → π 3 √ 3 − Tan X π − 3 X

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प्रश्न

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]

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उत्तर

\[\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{3} - \tan x}{\pi - 3x}\]
\[ = \lim_{h \to 0} \frac{\sqrt{3} - \tan \left( \frac{\pi}{3} - h \right)}{\pi - 3\left( \frac{\pi}{3} - h \right)}\]
\[ = \lim_{h \to 0} \frac{\sqrt{3} - \left( \frac{\tan \frac{\pi}{3} - \tan h}{1 + \tan \frac{\pi}{3} \tan h} \right)}{\pi - 3\left( \frac{\pi}{3} - h \right)}\]
\[ = \lim_{h \to 0} \frac{\sqrt{3} - \left( \frac{\sqrt{3} - \tan h}{1 + \sqrt{3} \tan h} \right)}{3h}\]
\[ = \lim_{h \to 0} \frac{\sqrt{3} + 3 \tan h - \sqrt{3} + \tan h}{\left( 1 + \sqrt{3} \tan h \right) 3h}\]
\[ = \lim_{h \to 0} \frac{4 \tan h}{3h \left( 1 + \sqrt{3}\tan h \right)}\]
\[ = \frac{4}{3}\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 8 | पृष्ठ ६२

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