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प्रश्न
\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x} \right]\] Dividing the numerator and the denominator by x:
\[\lim_{x \to 0} \left[ \frac{5\cos x + 3\left( \frac{\sin x}{x} \right)}{3x + \left( \frac{\tan x}{x} \right)} \right]\]
\[ = \left[ \frac{5\cos0 + 3}{3 \times 0 + 1} \right] \left[ \because \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1 \right]\]
\[ = \frac{5 + 3}{0 + 1}\]
\[ = 8\]
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