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Lim X → π 4 1 − Tan X X − π 4 - Mathematics

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प्रश्न

\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\] 

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उत्तर

\[\lim_{x \to \frac{\pi}{4}} \frac{1 - \tan x}{x - \frac{\pi}{4}}\]
\[ = \lim_{h \to 0} \frac{1 - \tan \left( \frac{\pi}{4} + h \right)}{\left( \frac{\pi}{4} + h - \frac{\pi}{4} \right)}\]
\[ = \lim_{h \to 0} \frac{1 - \left( \frac{1 + \tan h}{1 - \tan h} \right)}{h} \left[ \because \tan \left( \frac{\pi}{4} + 0 \right) = \frac{1 + \tan \theta}{1 - \tan \theta} \right]\]
\[ = \lim_{h \to 0} \frac{1 - \tan h - 1 - \tan h}{\left( 1 - \tan h \right) h}\]
\[ = \lim_{h \to 0} \frac{- 2 \tan h}{h\left( 1 - \tan h \right)} \left[ \because \lim_{h \to 0} \frac{\tan h}{h} = 1 \right]\]
\[ \Rightarrow \frac{- 2}{1 - 0} = - 2\]

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अध्याय 29: Limits - Exercise 29.8 [पृष्ठ ६२]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.8 | Q 6 | पृष्ठ ६२

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