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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Evaluate the following limits \[\] $ \lim\limits_{ x \to 0} \large\frac{ ax + x\cos\: x}{b\sin \: x}$

$\begin{array}{1 1}\frac{a+1}{b} \\ \frac{a}{b} \\ \frac{a+1}{b+1} \\ \frac{a}{1+b}\end{array} $

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1 Answer

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At $ x = 0 $ the value of the given function is $ \large\frac{0}{0}.$
Now,
$ \lim\limits_{x \to 0} \large\frac{ax + x \cos\: x}{b \sin \: x}$$ = \large\frac{1}{b}$$ \lim\limits_{x \to 0} \large\frac{x( a+ \cos \: x)}{\sin \: x}$
$ = \large\frac{1}{b}$$ \lim\limits_{ x \to 0} \bigg( \large\frac{x}{ \sin \: x} \bigg)$ $ \times \lim\limits_{x \to 0}$$( a + \cos\: x)$
$ = \large\frac{1}{b}$$ \times \large\frac{1}{\bigg( \lim\limits_{x \to 0} \Large\frac{\sin\: x}{x} \bigg)}$$ \times \lim\limits_{x \to 0} ( a + \cos\: x)$
$ \large\frac{1}{b}$$ \times ( a+ \cos \: 0) \quad \quad \bigg[ \lim\limits_{x \to 0} \large\frac{ \sin \: x}{x}$$ = 1 \bigg]$
$ = \large\frac{a+1}{b}$
answered Apr 4, 2014 by thanvigandhi_1
edited May 16, 2014 by balaji.thirumalai
 

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