Just for practice!#2

Calculus Level 4

Find a + b + c a+b+c such that

lim x 0 a x e x b . log ( 1 + x ) + c x e x x 2 s i n x = 2 \lim _{ x\rightarrow 0 }{ \frac { ax{ e }^{ x }-b.\log { (1+x) } +cx{ e }^{ -x } }{ { x }^{ 2 }sinx } } =\quad 2 .


The answer is 24.

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1 solution

Tanishq Varshney
Jul 5, 2015

x ( a e x + c e x ) b ln ( 1 + x ) x 2 sin x \large{\frac{x(ae^x+ce^{-x})-b\ln (1+x)}{x^2 \sin x}}

using series expansion of e x , e x , sin x , ln ( 1 + x ) e^{x},e^{-x},\sin x ,\ln (1+x)

x ( a + a x + a x 2 2 + . . + c c x + c x 2 2 . . . ) b ( x x 2 2 + x 3 3 . . . ) x 2 ( x x 3 3 ! + . . . ) \large{\frac{x(a+ax+\frac{ax^2}{2}+..+c-cx+\frac{cx^2}{2}-...)-b(x-\frac{x^2}{2}+\frac{x^{3}}{3}-...)}{x^2 (x-\frac{x^{3}}{3!}+...)}}

collecting terms of x , x 2 a n d x 3 x,x^{2}~and ~x^3

x ( a b + c ) + x 2 ( a c + b 2 ) + x 3 ( a 2 + c 2 b 3 ) x 3 ( 1 x 2 3 ! + . . ) \large{\frac{x(a-b+c)+x^{2}(a-c+\frac{b}{2})+x^{3}(\frac{a}{2}+\frac{c}{2}-\frac{b}{3})}{x^{3} (1-\frac{x^2}{3!}+..)}}

For limit to exist

{ a b + c = 0 a c + b 2 = 0 a 2 + c 2 b 3 = 2 \large{\begin{cases} a-b+c=0 \\ a-c+\frac { b }{ 2 } =0 \\ \frac { a }{ 2 } +\frac { c }{ 2 } -\frac { b }{ 3 } =2 \end{cases}}

solving we get a = 9 b = 12 c = 3 \boxed{a=9}\quad \boxed{b=12}\quad \boxed{c=3}

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