Find the values of $p$ and $q$ such that $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13.$ Assume, that L' Hospital's rule is applicable.

Question

Find the values of $p$ and $q$ such that $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13.$ Assume, that L' Hospital's rule is applicable.

I tried solving the problem as follows: Given, $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\frac 13$ and according to the given information, that L' Hospital's rule is applicable we have, $\lim_{x\to 0}\frac{x(1-p\cos x)+q\sin x}{x^3}=\lim_{x\to 0}\frac{(1-p\cos x)+xp\sin x+q\cos x}{3x^2}=\frac 13.$ Applying L' Hospital's rule again, we have, $\lim_{x\to 0}\frac{(1-p\cos x)+xp\sin x+q\cos x}{3x^2}=\lim_{x\to 0}\frac{p\sin x+xp\cos x+p\sin x-q\sin x}{6x}=\frac 13\implies 3p-q=2.$

This means $p=q=1$ is a solution of the equation.

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However, the answer given in the book is, $p=\frac 12=-q.$ However, I don't really understand where did I make the mistake?

Is the way I tried solving this problem a valid approach?

Answer 1

After you get

$\lim_{x\to 0}\frac{(1-p\cos x)+xp\sin x+q\cos x}{3x^2}$,

it can be written as

$\lim_{x\to 0}\frac{1-(p-q)\cos x+xp\sin x}{3x^2}$.

At this moment, it should be noted that $p-q$ must be $1$, otherwise the limit will be $\infty$ or $-\infty$. Thus, we have $p-q=1$, and

$\lim_{x\to 0}\frac{1-\cos x+xp\sin x}{3x^2}=\lim_{x\to 0}\frac{\sin x+p(\sin x+x\cos x)}{6x}=\lim_{x\to 0}\frac{\cos x+p(\cos x+\cos x-x\sin x)}{6}$ $=\frac{1+2p}{6}$ .

Now letting $\frac{1+2p}{6}=\frac{1}{3}$ yields $p=\frac{1}{2}$ and then $q = -\frac{1}{2}$.

Answer 2

I would solve the exercise as follows. It is convenient to transform the numerator so as the expressions $1-\cos x$ and $\sin x-x$ show up there. We have $${x(1-p\cos x)+q\sin x\over x^3}={1+q-p\over x^2}+p{1-\cos x\over x^2}+q{\sin x-x\over x^3}$$ Then $$\lim_{x\to 0}{1-\cos x\over x^2}\overset{(H)}{=}{1\over 2}\lim_{x\to 0}{\sin x\over x}={1\over 2}$$ $$\lim_{x\to 0}{\sin x-x\over x^3}\overset{(H)}{=}\lim_{x\to 0}{\cos x-1\over 3x^2}=-{1\over 6}$$ The limit of the first term exists (and is equal $0$) only for $p=q+1.$ The result is equal ${1\over 2}p-{1\over 6}q.$ Solving the system of two equation yields $p={1\over 2}=-q.$