$f(x)$ is a function such that $$\lim_{x\to0} \frac{f(x)}{x}=1$$ if
$$\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}=1$$
Find $a$ and $b$
Can I assume $f(x)$ to be $\sin(x)$ since $\sin$ satisfies the given condition?
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Hint. You may use the standard Taylor series expansions, as $x \to 0$, $$ \begin{align} \cos x&=1-\frac{x^2}2+O(x^4)\\ \sin x&=x-\frac{x^3}6+O(x^4) \end{align} $$ giving $$ \begin{align} \frac{x(1+a\cos x)-b\sin x}{(f(x))^3}&=\frac{(1+a-b) x+\frac16 (-3 a+b) x^3+O(x^5)}{(f(x))^3} \\\\&=\frac{(1+a-b) x+\frac169 (-3 a+b) x^3+O(x^5)}{x^3(1+\epsilon(x))^3} \end{align} $$ where, as $x \to 0$, we have used $f(x)=x(1+\epsilon(x))$ with $\epsilon(x) \to 0$.
Can you take it from here?
Olivier Oloa
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@almagest Sorry, I did not read correctly the question. Let's say it is a hint to answer the first part of the question: "find $a$ and $b$..." – Olivier Oloa Apr 22 '16 at 07:52
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To vanish the first term in $x$ gives $b=a+1$, then you have to solve $\frac{1}{6} (-3 a+b)=1$, you find $a=-\frac52$. – Olivier Oloa Apr 22 '16 at 08:09
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If not the limit doesn't exist, if not you get something like $\frac{(1+a-b)x}{x^3}$ which is bad as $x \to 0$... – Olivier Oloa Apr 22 '16 at 08:14
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Not exactly, we have $f(x)=x+x\epsilon(x)$, that is $x$ plus something small before $x$... because $\lim f(x)/x=1$ means $f(x)/x-1=\epsilon(x)$ with $\epsilon(x) \to 0$, then multiplying throughout by $x$ gives $f(x)=x+x\epsilon(x)$. – Olivier Oloa Apr 22 '16 at 08:30
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Rewrite $$A=\frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}= \frac{x(1+a\cos(x))-b\sin(x)}{x^3}\times\Big(\frac{x}{f(x)}\Big)^3$$ and now, as Olivier Oloa answered, use Taylor expansions.
Whichever could be $\lim_{x\to0} \frac{f(x)}{x}=L$ (except $0$ or $\infty$), you could find the conditions you want for $\lim_{x\to0} A=M$.
Claude Leibovici
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Hint: we have: $\dfrac{1+a\cos x}{x^2} - \dfrac{b}{\sin^2 x} \to 1$, and rewrite $\dfrac{b}{\sin^2 x} = \dfrac{b}{x^2}\cdot \dfrac{x^2}{\sin^2 x}$ then the limit on the left equals $\dfrac{1+a\cos x - b}{x^2} = 1$, this means you can assume L'hopitale rule meaning: $1 + a - b = 0$, and differentiate both numerator and denominator to get another equation and solve for $a, b$.
DeepSea
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Since $\displaystyle\lim_{x\to0}\frac{f(x)}{x}=1$, we have $$\lim_{x\to0}\frac{x}{f(x)} =\lim_{x\to0}\frac{1}{\frac{f(x)}{x}} =\frac{1}{\displaystyle\lim_{x\to0}\frac{f(x)}{x}} =\frac{1}{1}=1.$$ Now, by using L'Hopital's rule, we have \begin{align} 1&=\lim_{x \to 0} \frac{x(1+a\cos(x))-b\sin(x)}{f(x)^3}\\ &=\lim_{x \to 0}\left[\frac{x^3}{f(x)^3}\cdot\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\right]\\ &=\left(\lim_{x \to 0}\frac{x}{f(x)}\right)^3\cdot \lim_{x\to0}\frac{x(1+a\cos(x))-b\sin(x)}{x^3}\\ &\stackrel{\rm H}{=} 1^3\cdot\lim_{x\to0}\frac{1+a\cos(x)-ax\sin(x)-b\cos(x)}{3x^2}\tag{1}\\ &\stackrel{\rm H}{=} \lim_{x\to0}\frac{-a\sin(x)-a\sin(x)-ax\cos(x)+b\sin(x)}{6x}\\ &\stackrel{\rm H}{=} \lim_{x\to0}\frac{-2a\cos(x)-a\cos(x)+ax\sin(x)+b\cos(x)}{6}\\ &=\frac{-3a+b}{6}\tag{2}. \end{align} By $(1)$, we have to force the limit of the numerator to be zero, so $1+a-b=0$. Also, by $(2)$ we naturally get $-3a+b=6$. Hence we conclude that $a=-\frac{5}{2}$ and $b=-\frac{3}{2}$.
Solumilkyu
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