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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.

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  • $\begingroup$ Break into fractions and then you can see what you can do. $\endgroup$
    – jayant98
    Commented Dec 7, 2018 at 14:54
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    $\begingroup$ Can you use Taylor expansions? $\endgroup$
    – MSDG
    Commented Dec 7, 2018 at 14:59
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    $\begingroup$ I believe you are missing a term in the numerator. $\endgroup$ Commented Dec 7, 2018 at 15:01
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    $\begingroup$ Indeed, right now this doesn't seem to have a solution. $\endgroup$
    – RcnSc
    Commented Dec 7, 2018 at 15:03
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    $\begingroup$ 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. $\endgroup$ Commented Dec 7, 2018 at 15:12

3 Answers 3

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

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    $\begingroup$ 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). $\endgroup$
    – MSDG
    Commented Dec 7, 2018 at 15:23
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    $\begingroup$ @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.) $\endgroup$
    – J.G.
    Commented Dec 7, 2018 at 15:24
  • $\begingroup$ Good point! Seems that I have forgotten how these standard limits are actually proven! $\endgroup$
    – MSDG
    Commented Dec 7, 2018 at 15:26
  • $\begingroup$ I think the right term should be "no derivatives". I consider limits to be very much a part of calculus. +1 $\endgroup$
    – Paramanand Singh
    Commented Dec 8, 2018 at 2:17
  • $\begingroup$ @ParamanandSingh I see them more as analysis. $\endgroup$
    – J.G.
    Commented Dec 8, 2018 at 7:43
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$$f(x)=\frac{p + q\cos x + r\sin x}{x^2}$$.

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

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

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

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

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  • $\begingroup$ The use of L'Hospital's Rule in reverse is not allowed. You may want to fix it. $\endgroup$
    – Paramanand Singh
    Commented Dec 8, 2018 at 2:18
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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 $p=1,q=-1,r=0$.

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