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Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.

My attempt:
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$

$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$

$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$

$\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$

$\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$

As denominator is tending to zero,so numerato will also tend to zero.

$1+a-b=0............................(1)$

Applying L Hospital rule,
$\lim_{x\to 0}\frac{-a\sin x}{2x}=1$
So $a=-2$ and $b=-1$

But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$.

I do not understand where have i gone wrong?

Did
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Brahmagupta
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    Transition from the 4th to the 5th line is incorrect. You can't pass to a limit inside a limit like that. – Wojowu Mar 26 '16 at 08:11

3 Answers3

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Since we can write $$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\cdot\frac{1}{(f(x)/x)^3}$$ we have to have $$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$$ We can write $$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\tag1$$$$=\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}\tag2$$ But we cannot write $(2)$ as $$\lim_{x\to 0}\frac{1+a\cos x-b\cdot 1}{x^2}$$

From $(1)$, by L'Hôpital's rule, $$(1)=\lim_{x\to 0}\frac{1+a\cos x-ax\sin x-b\cos x}{3x^2}\tag3$$ Here, we have to have $$1+a-b=0\tag 4$$ Using L'Hôpital's rule several times, $$\begin{align}(3)&=\lim_{x\to 0}\frac{1-ax\sin x-\cos x}{3x^2}\\&=\lim_{x\to 0}\frac{-a(\sin x+x\cos x)+\sin x}{6x}\\&=\lim_{x\to 0}\frac{-2a\cos x+ax\sin x+\cos x}{6}\\&=\frac{-2a+1}{6}\end{align}$$ and so $$\frac{-2a+1}{6}=1\tag5$$ Now solve $(4)(5)$.

mathlove
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  • I do not understand why cant we replace $\frac{\sin x}{x}$ by $1$ inside the expression,as i think as far as limits exists,we can plug the standard limits into complex expressions of limits.Am i wrong?@mathlove – Brahmagupta Mar 26 '16 at 08:57
  • @Brahmagupta: Basically, the error is similar to the following : since we have $\lim_{n\to\infty}1/n=0$, we have $\lim_{n\to\infty}(1+1/n)^n=\lim_{n\to\infty}(1+0)^n=\lim_{n\to\infty}1^n=1$, which is wrong. I hope this helps. – mathlove Mar 26 '16 at 11:28
  • @Brahmagupta: Replacing a sub-expression with its limit while calculating the limit of an expression is allowed only in two scenarios 1) when the sub-expression is connected to rest of expression in additive manner and 2) when sub-expression is connected to rest of the expression in multiplicative manner and limit of sub-expression is non-zero. See http://math.stackexchange.com/a/1659261/72031 – Paramanand Singh Mar 27 '16 at 06:29
  • @ParamanandSingh,in my question,the sub-expression $\frac{\sin x}{x}$is connected to rest of the expression in additive manner but it is making calculations faulty when i replace its value by $1$ – Brahmagupta Mar 27 '16 at 06:53
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    @Brahmagupta: No. Let the complicated expression be called $C$ and sub-expression be called $S$. If we remove $S$ from $C$ we get rest of expression say $R$. For additive connection we must have $$C=R\pm S$$ Here $$S=\frac{b\sin x}{x},C=\dfrac{1+a\cos x -\dfrac{b\sin x}{x}}{x^{2}}$$ and $$R=\frac{1+a\cos x}{x^{2}}$$ and clearly we have neither $C=R+S$ nor $C=R-S$. – Paramanand Singh Mar 27 '16 at 07:13
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Just as mathlove answered, we need to find $a,b$ such that $$\lim_{x\to 0}\frac{x(1+a\cos (x))-b\sin (x)}{x^3}=1$$ Let us use the classical Taylor expansions $$\cos(x)=1-\frac{x^2}{2}+O\left(x^3\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ and replace to get $$\frac{x(1+a\cos (x))-b\sin (x)}{x^3}=\frac{x (a-b+1)+x^3 \left(\frac{b}{6}-\frac{a}{2}\right)+O\left(x^4\right)}{x^3}$$ So, the first thing is $a-b+1=0$ and the second (for a limit equal to $1$) is $\frac{b}{6}-\frac{a}{2}=1$. Solving these two equations leads to the required values.

Claude Leibovici
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1

Others have pointed to your mistake.

The numerator in your third line can be written as $$x\bigl(1+a(1-{x^2\over2}+?x^4)\bigr)-b\bigl(x-{x^3\over6}+?x^5\bigr)=(1+a-b)x+\bigl(-{a\over2}+{b\over6}\bigr)x^3+?x^5\ ,$$ where each question mark represents some convergent power series. It follows that $a$ and $b$ have to satisfy $$1-a-b==0,\qquad-{a\over2}+{b\over6}=1\ ,$$ from which we obtain $$a=-{5\over2},\qquad b=-{3\over2}\ .$$

Christian Blatter
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