For what $p,q,r$ $\lim_{x \to 0} f(x)=\frac{p + q\cos x + r\sin x}{x^2}=1/2$? Let $$f(x)=\frac{p + q\cos x + r\sin x}{x^2}$$. Then for what values of $p$,$q$ and $r$ is the limit $$\lim_{x \to 0} f(x)=1/2.$$ I’ve tried using L’Hospital’s rule but couldn’t get anywhere. Any help would be appreciated.
 A: No calculus is needed. Since $\frac{\sin x}{x}\to 1$, $\frac{\cos x-1}{x^2}=-\frac{2\sin^2 x/2}{x^2}\to-\frac{1}{2}$. But $\frac{1}{x^2},\,\frac{\sin x}{x^2}$ diverge, so take $p=1,\,q=-1,\,r=0$.
A: $$f(x)=\frac{p + q\cos x + r\sin x}{x^2}$$.
$$\lim_{x\rightarrow 0}f(x)=\frac{1}{2} \implies p+q=0$$
$$ \lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow 0} \frac{-q\sin x+r\cos  x}{2x}=\frac{1}{2} \implies r=0$$
$$\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{-q\cos x -r\sin x}{2}=\frac{1}{2}\implies -q=1 \implies q=-1.$$
Therefore, $p=1, p=-1,$ and $r=0$.
A: We are given that $$\lim_{x\to 0}\frac{p+q\cos x+r\sin x} {x^2}=\frac{1}{2}\tag{1}$$ so that multiplication by $x^2$ gives $$\lim_{x\to 0}(p+q\cos x+r\sin x) =\lim_{x\to 0}\frac{p+q\cos x+r\sin x} {x^2}\cdot x^2=\frac{1}{2}\cdot 0=0$$ and hence $p+q=0$.
Next note that $$p+q\cos x+r\sin x=p+q-q(1-\cos x) +r\sin x=r\sin x-q(1-\cos x) $$ and since $(1-\cos x) /x^2\to 1/2$ it follows from $(1)$ that $$\lim_{x\to 0}\frac{r\sin x} {x^2}=\frac{1+q}{2}$$ and since $(\sin x) /x\to 1$ we have $$\lim_{x\to 0}\frac{r}{x}=\frac{1+q}{2}\tag{2}$$ Multiplying by $x$ gives us $r=0$ and then from the above equation we get $1+q=0$ so that $p=1,q=-1,r=0$.
