MCQ
If $\mathop {Lim}\limits_{x\, \to \,0} \, (x^{-3} sin 3x + ax^{-2} + b)$ exists and is equal to zero then :
- ✓$a = - 3\, \& \,b = 9/2$
- B$a = 3\, \& \,b = 9/2$
- C$a = - 3\, \& \,b = -9/2$
- D$a = 3\, \& \,b = -9/2$
$=\mathop {Lim}\limits_{x\, \to \,0} $ $\frac{{\sin 3x\, + ax + b{x^3}}}{{{x^3}}}\,$
$= \mathop {Lim}\limits_{x\, \to \,0} $ $\frac{{3\,\frac{{\sin 3x}}{{3x}}\,\, + \,\,a\,\, + \,\,b{x^2}}}{{{x^2}}}\,$
for existence of limit $3 + a = 0 \Rightarrow a = - 3 $
$\therefore l =$$\mathop {Lim}\limits_{x\, \to \,0} $$\frac{{\sin 3x\, - 3x\, + b{x^3}}}{{{x^3}}}\,$$= 27\,.\,\frac{{\sin t - t}}{{{t^3}}}\, + \,b\,=0$ $ (3x = t)$
$= - \,\frac{{27}}{6}\,\, + \,\,b\,\, = 0 \Rightarrow b = \frac{9}{2}\,$
[ $OR$ use $L$’ Hospital’s rule ]
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