Question

Solve for a and b: $\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$.

Answer 1

We have $\underset{x->0}{lim}\frac{\left[x\right(1+a\cos x)-b\sin x]}{x^3}=\frac{1}{3}$

but $\underset{x->0}{lim}\frac{\left[x\right(1+a\cos x)-b\sin x]}{x^3}$ has the

form $\frac{0}{0}$ so from L' Hospital's Rule

taking derivatives we have $\underset{x->0}{lim}\frac{[1+a\cos x-ax\sin x-b\cos x]}{x^3}=\frac{1}{3}$ ---(1)

Putting $g\left(x\right)=\frac{[1-a\cos x+ax\sin x+b\cos x]}{3x^2}$

$\left(3x^2\right)g\left(x\right)=[1-a\cos x+ax\sin x+b\cos x]$

Taking $\underset{x->0}{lim}$ of both sides we have

$0\ast \frac{1}{3}=1-a+b$ $⟹b=a-1$ ---(2)

(1) because of (2) becomes

$\underset{x->0}{lim}\frac{[1+a\cos x-ax\sin x-a\cos x+\cos x]}{3x^2}=\frac{1}{3}$

$\underset{x->0}{lim}\frac{[1-ax\sin x+\cos x]}{3x^2}=\frac{1}{3}$

L'Hospital's rule again

$\underset{x->0}{lim}\frac{[-a\sin x-ax\cos x-\sin x]}{6x}=\frac{1}{3}$

L'Hospital's rule again

$\underset{x->0}{lim}\frac{[-a\cos x-a\cos x+ax\sin x-\cos x]}{6}=\frac{1}{3}$

$\underset{x->0}{lim}\frac{[-2a\cos x+ax\sin x-\cos x]}{6}=\frac{1}{3}$

Should be

$\underset{x->0}{lim}[-2a\cos x+ax\sin x-\cos x]=2$

$-2a-1=2$ $⟹a=\frac{-3}{2}andb=\frac{-5}{2}$