If f(x)=sin3x+Asin2x+Bsinxx5, x≠0, continuous at x=0, then A+B+f(0)= --------

f(0)=limx→0sin3x+Asin2x+Bsinxx500 form using l'hospital's rulef(0)=Ltx→03cos3x+2Acos2x+Bcosx5x4=3+2A+B0 Exists⇒2A+B=−3f(0)=limx→0-9sin3x-4Asin2x-Bsinx20X3 00 form using L'Hospital rulef(0)=Ltx→0−27cos3x−8Acos2x−Bcosx60x2=-27-8A-B0Exists 8A+B=−27⇒A=−4,B=5f(0)=limx→081sin3x+16Asin2x+Bsinx120x 00 form using L'Hospital rule =limx→081sin3xx+16Asin2xx+Bsinxx120 =81.3+32(-4)+5120=243-128+5120=1 ∴A+B+f(0)=−4+5+1=2

If f(x)=sin3x+Asin2x+Bsinxx5, x≠0, continuous at x=0, then A+B+f(0)= --------

f(0)=limx→0sin3x+Asin2x+Bsinxx500 form using l'hospital's rulef(0)=Ltx→03cos3x+2Acos2x+Bcosx5x4=3+2A+B0 Exists⇒2A+B=−3f(0)=limx→0-9sin3x-4Asin2x-Bsinx20X3 00 form using L'Hospital rulef(0)=Ltx→0−27cos3x−8Acos2x−Bcosx60x2=-27-8A-B0Exists 8A+B=−27⇒A=−4,B=5f(0)=limx→081sin3x+16Asin2x+Bsinx120x 00 form using L'Hospital rule =limx→081sin3xx+16Asin2xx+Bsinxx120 =81.3+32(-4)+5120=243-128+5120=1 ∴A+B+f(0)=−4+5+1=2

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If $f (x) = \frac{\sin 3 x + A \sin 2 x + B \sin x}{x^{5}}, x \neq 0,$ continuous at $x = 0,$ then $A + B + f (0) =$ --------

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Detailed Solution

$f (0) = \lim_{x \to 0} \frac{\sin 3 x + A \sin 2 x + B \sin x}{x^{5}} (\frac{0}{0} f o r m)$ using l'hospital's rule

$f (0) = \underset{x \to 0}{L t} \frac{3 \cos 3 x + 2 A \cos 2 x + B \cos x}{5 x^{4}} = \frac{3 + 2 A + B}{0}$

Exists $\Rightarrow 2 A + B = - 3$

$f (0) = \lim_{x \to 0} \frac{- 9 \sin 3 x - 4 A \sin 2 x - B \sin x}{20 X^{3}} (\frac{0}{0} f o r m u \sin g L' H o s p i t a l r u l e)$

$f (0) = \underset{x \to 0}{L t} \frac{- 27 \cos 3 x - 8 A \cos 2 x - B \cos x}{60 x^{2}} = \frac{- 27 - 8 A - B}{0}$

Exists $8 A + B = - 27$

$\Rightarrow A = - 4, B = 5$

$f (0) = \lim_{x \to 0} \frac{81 \sin 3 x + 16 A \sin 2 x + B \sin x}{120 x} (\frac{0}{0} f o r m u \sin g L' H o s p i t a l r u l e) = \lim_{x \to 0} \frac{81 \frac{\sin 3 x}{x} + 16 A \frac{\sin 2 x}{x} + B \frac{\sin x}{x}}{120} = \frac{81.3 + 32 (- 4) + 5}{120} = \frac{243 - 128 + 5}{120} = 1$