Question

If $Li{m}_{x\to 0}\frac{4+sin2x+Asinx+Bcosx}{x^2}$ exists, then the values A and B are

• A. -2 and -4
• B. -4 and -2
• C. -3 and -2
• D. -1 and -4

Answer 1

The correct option is A

-2 and -4

Since the given limit exists and denominator approaches zero as x $\to 0$

Numerator must approach zero as x $\to 0$ that is possible when 4 + B = 0 which is obtained by substituting zero in numerator in place of x.

$\therefore B=-4\therefore {lim}_{x\to 0}\frac{4+sin2x+Asinx-4cosx}{x^2}$ [0/0 form]

$\therefore {lim}_{x\to 0}\frac{2cos2x+Acosx+4sinx}{2x}$ [By L'Hospital Rule]

Again for above limit to exist Numerator must approach to zero since denominator is also approaching to zero.
So, A + 2 =0, $\Rightarrow$ A = -2

$\therefore {lim}_{x\to 0}\frac{2cos2x-2cosx+4sinx}{2x}\left[0/0form\right]\therefore {lim}_{x\to 0}\frac{-4sin2x+2sinx+4cosx}{2}=\frac{4}{2}=2\therefore A=-2andB=-4$