# 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.

• Break into fractions and then you can see what you can do. – jayant98 Dec 7 at 14:54
• Can you use Taylor expansions? – MisterRiemann Dec 7 at 14:59
• I believe you are missing a term in the numerator. – user10354138 Dec 7 at 15:01
• Indeed, right now this doesn't seem to have a solution. – RcnSc Dec 7 at 15:03
• Hint (as @MisterRiemann suggests). Write out the first few terms of the power series for $\sin$ and $\cos$, do the algebra and see what you can conclude. L'Hopital is almost always a bad first tool for a job like this. – Ethan Bolker Dec 7 at 15:12

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 $$p=1,\,q=-1,\,r=0$$.

• Well, you're using standard limits, so I don't think it's fair to say that no calculus is needed. :-) Nice solution otherwise (+1). – MisterRiemann Dec 7 at 15:23
• @MisterRiemann No, these limits are famously provable without calculus. I just proved one from the other, whereas $\frac{\sin x}{x}$ is the portion of an angle-$x$ sector contained in the triangle with the same vertices. (That's not the whole proof, but I'll leave you to look up the rest.) – J.G. Dec 7 at 15:24
• Good point! Seems that I have forgotten how these standard limits are actually proven! – MisterRiemann Dec 7 at 15:26
• I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1 – Paramanand Singh Dec 8 at 2:17
• @ParamanandSingh I see them more as analysis. – J.G. Dec 8 at 7:43

$$f(x)=\frac{p + q\cos x + r\sin x}{x^2}$$.

$$\lim_{x\rightarrow 0}f(x)=\frac{1}{2} \implies p+q=0$$

$$\lim_{x\rightarrow 0}f(x)= \lim_{x\rightarrow 0} \frac{-q\sin x+r\cos x}{2x}=\frac{1}{2} \implies r=0$$

$$\lim_{x\rightarrow 0}f(x)=\lim_{x\rightarrow 0}\frac{-q\cos x -r\sin x}{2}=\frac{1}{2}\implies -q=1 \implies q=-1.$$

Therefore, $$p=1, p=-1,$$ and $$r=0$$.

• The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it. – Paramanand Singh Dec 8 at 2:18

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 $$p=1,q=-1,r=0$$.