If f(x)=asinx−bx+cx2+x32x2ℓn(1+x)−2x3+x4, when x≠0 and f(x) is continuous at x=0, find value of 200×f(0)
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Solution
limx→0a[x−x36+x5120−.......]−bx+cx2+x32x2[x−x22+x33.........]−2x3+x4 =limx→0(a−b)x+cx2+[1−a6]x3+ax5120+.....2x53−x62+...... for this limit to be exist, a−b=0,c=0, & 1−a6=0 ⇒a=b=6 & c=0. then f(0)=6120.32=340⇒200f(0)=15