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LIMITS AND DERIVATIVES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSLIMITS AND DERIVATIVES2 Marks
Question
Evaluate $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}};$ a, b, a + b $\neq$ 0.
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Answer

We have,
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax + bx}}{{ax + \sin bx}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{\sin ax}}{x} + \frac{{bx}}{x}}}{{\frac{{ax}}{x} + \frac{{\sin bx}}{x}}}$
[dividing both numerator and denominator by x]
$ = \frac{{\mathop {\lim }\limits_{x \to 0} \frac{{\sin (ax)}}{{ax}} \times a + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + \mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}} \times b}}$$\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}} = \frac{{\mathop {\lim }\limits_{x \to a} f(x)}}{{\mathop {\lim }\limits_{x \to a} g(x)}}{\text{ and }}} {\mathop {\lim }\limits_{x \to a} f(x) + g(x) = \mathop {\lim }\limits_{x \to a} f(x) + \mathop {\lim }\limits_{x \to a} g(x)} \right]$
$ = \frac{{a\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{ax}} + \mathop {\lim }\limits_{x \to 0} b}}{{\mathop {\lim }\limits_{x \to 0} a + b\mathop {\lim }\limits_{x \to 0} \frac{{\sin bx}}{{bx}}}}$
$= \frac { a ( 1 ) + b } { a + b ( 1 ) }$ $\left[ {\because \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1} \right]$
$= \frac { a + b } { a + b } = 1$

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