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Long back I had encountered the following problem in Hardy's Pure Mathematics (originally from the infamous Mathematical Tripos 1896):

If $$f(x) = \frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a}$$ then show that $$\frac{d}{da}\{\lim_{x \to a}f(x)\} - \lim_{x \to a}f'(x) = \frac{3}{4}\sec^{3}a - \frac{5}{12}\sec a$$

I had solved it using Taylor series expansions and even then it involved good amount of calculation. I suppose applying L'Hospital would be even more arduous.

On the other hand both the Taylor series and L'Hospital Rule are discussed later in Hardy's book suggesting that it could be solved via elementary techniques (i.e. using algebra of limits and Squeeze theorem and if needed one can use mean value theorem). Please let me know if such a solution is possible.

Also I believe there might be a suitable generalization applicable to functions of type $$g(x) = \frac{1}{\phi(x) - \phi(a)} - \frac{1}{(x - a)\phi'(a)}$$ and perhaps the expression $$\frac{d}{da}\{\lim_{x \to a}g(x)\} - \lim_{x \to a}g'(x)$$ has some significance. Any ideas in this direction would be helpful.

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This is NOT a complete answer but just a small contribution to it. Please consider it as a hint. First, recall that \begin{equation*} f(x)=\frac{1}{\sin x-\sin a}-\frac{1}{(x-a)\cos a} \end{equation*} We have to prove that \begin{equation*} \frac{d}{da}\{\lim_{x\rightarrow a}f(x)\}-\lim_{x\rightarrow a}f^{\prime }(x)=\frac{3}{4}\sec ^{3}a-\frac{5}{12}\sec a. \end{equation*} This is the asked original problem. Since it seems to be difficult then to solve it, I suggests to split it into four 'small' sub-problems as follows.

  1. $\bf{[P1.]}$ Show that \begin{equation*} f^{\prime }(x)=\frac{\sec a}{(x-a)^{2}}-\frac{\cos x}{(\sin x-\sin a)^{2}}. \end{equation*}
  2. $\bf{[P2.]}$ Show that \begin{equation*} \lim_{x\rightarrow a}f^{\prime }(x)=\frac{\sec ^{3}a}{4}-\frac{\sec a}{12}. \end{equation*}
  3. $\bf{[P3.]}$ Show that \begin{equation*} \lim_{x\rightarrow a}f(x)=\frac{1}{2}\sec a\tan a. \end{equation*}
  4. $\bf{[P4.]}$ Show that \begin{equation*} \frac{d}{da}\{\lim_{x\rightarrow a}f(x)\}=\sec ^{3}a-\frac{1}{2}\sec a. \end{equation*} In fact, if these sub-problems are solved, then the original problem follows \begin{equation*} \frac{d}{da}\{\lim_{x\rightarrow a}f(x)\}-\lim_{x\rightarrow a}f^{\prime }(x)=\left( \sec ^{3}a-\frac{1}{2}\sec a\right) -\left( \frac{\sec ^{3}a}{4}-% \frac{\sec a}{12}\right) =\frac{3}{4}\sec ^{3}a-\frac{5}{12}\sec a. \end{equation*}

The assertion $\bf{[P1.]}$ follows after simple computations.

If we assume $\bf{[P3.]}$ then $\bf{[P4.]}$ follows easily since \begin{equation*} \frac{d}{da}\{\lim_{x\rightarrow a}f(x)\}=\frac{d}{da}\{\frac{1}{2}\sec a\tan a\}=\sec ^{3}a-\frac{1}{2}\sec a\ \ \ \ \text{after simple computations.} \end{equation*} Therefore the original problem is converted into the two problems $\bf{[P2.]}$ and $\bf{[P3.]}$ cited above.

$\bf{UPDATE1:}$ I will solve P3, that is I will prove that \begin{equation*} \lim_{x\rightarrow a}\frac{1}{\sin x-\sin a}-\frac{1}{(x-a)\cos a}=\frac{1}{2% }\sec a\tan a. \end{equation*} I will use the'standard limits \begin{eqnarray*} \lim_{x\rightarrow a}\frac{\sin x-\sin a}{x-a} &=&\cos a \\ \lim_{x\rightarrow a}\frac{\sin x-\sin a-(x-a)\cos a}{\left( x-a\right) ^{2}} &=&-\frac{1}{2}\sin a. \end{eqnarray*} \begin{eqnarray*} \frac{1}{\sin x-\sin a}-\frac{1}{(x-a)\cos a} &=&\frac{(x-a)\cos a-\sin x+\sin a}{\left( \sin x-\sin a\right) (x-a)\cos a} \\ &=&-\frac{\frac{\sin x-\sin a-(x-a)\cos a}{(x-a)^{2}}}{\frac{\left( \sin x-\sin a\right) (x-a)\cos a}{(x-a)^{2}}} \\ &=&-\frac{\frac{\sin x-\sin a-(x-a)\cos a}{(x-a)^{2}}}{\frac{\left( \sin x-\sin a\right) }{(x-a)}\cos a} \end{eqnarray*} then \begin{equation*} \lim_{x\rightarrow a}\frac{1}{\sin x-\sin a}-\frac{1}{(x-a)\cos a}% =-\lim_{x\rightarrow a}\frac{\frac{\sin x-\sin a-(x-a)\cos a}{(x-a)^{2}}}{% \frac{\left( \sin x-\sin a\right) }{(x-a)}\cos a}=-\frac{-\frac{1}{2}\sin a}{% \cos a\cos a}=\frac{1}{2}\sec a\tan a. \end{equation*} $\bf{UPDATE2:}$ I\ will solve P2 as follows. \begin{equation*} \frac{\sec a}{(x-a)^{2}}-\frac{\cos x}{(\sin x-\sin a)^{2}}=\frac{\sec a(\sin x-\sin a)^{2}-\cos x(x-a)^{2}}{(x-a)^{2}(\sin x-\sin a)^{2}} \end{equation*}% \begin{equation*} =\frac{\sec a(\sin x-\sin a-\cos a(x-a)+\cos a(x-a))^{2}-(\cos x-\cos a+\sin a(x-a)+\cos a-\sin a(x-a))(x-a)^{2}}{(x-a)^{2}(\sin x-\sin a)^{2}} \end{equation*}

Consider the left part of the numerator \begin{eqnarray*} \sec a([\sin x-\sin a-\cos a(x-a)]+\cos a(x-a))^{2} &=&\sec a(\sin x-\sin a-\cos a(x-a))^{2} \\ &&+2(x-a)(\sin x-\sin a-\cos a(x-a)) \\ &&+\cos a(x-a)^{2} \end{eqnarray*} now the right part of the numerator% \begin{eqnarray*} (\cos x-\cos a+\sin a(x-a)+\cos a-\sin a(x-a))(x-a)^{2} &=&\left( \cos x-\cos a+\sin a(x-a)\right) (x-a)^{2} \\ &&+(\cos a)(x-a)^{2}-\sin a(x-a)^{3} \end{eqnarray*} now the full numerator \begin{eqnarray*} &&\sec a([\sin x-\sin a-\cos a(x-a)]+\cos a(x-a))^{2} \\ &&-(\cos x-\cos a+\sin a(x-a)+\cos a-\sin a(x-a))(x-a)^{2} \\ &=&\sec a(\sin x-\sin a-\cos a(x-a))^{2} \\ &&+2(x-a)(\sin x-\sin a-\cos a(x-a)) \\ &&+(\cos a)(x-a)^{2}-\left( \cos x-\cos a+\sin a(x-a)\right) (x-a)^{2}-(\cos a)(x-a)^{2}+\sin a(x-a)^{3} \end{eqnarray*} \begin{eqnarray*} &=&\sec a(\sin x-\sin a-\cos a(x-a))^{2} \\ &&+2(x-a)(\sin x-\sin a-\cos a(x-a)) \\ &&-\left( \cos x-\cos a+\sin a(x-a)\right) (x-a)^{2} \\ &&+\sin a(x-a)^{3} \end{eqnarray*} \begin{eqnarray*} &=&\sec a(\sin x-\sin a-\cos a(x-a))^{2} \\ &&-\left( \cos x-\cos a+\sin a(x-a)\right) (x-a)^{2} \\ &&+2(x-a)(\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}) \end{eqnarray*} Now We have \begin{eqnarray*} \lim_{x\rightarrow a}f^{\prime }(x) &=&\lim_{x\rightarrow a}\frac{\frac{\sec a(\sin x-\sin a)^{2}-\cos x(x-a)^{2}}{(x-a)^{4}}}{\frac{(x-a)^{2}(\sin x-\sin a)^{2}}{(x-a)^{4}}} \\ &=&\lim_{x\rightarrow a}\frac{\sec a(\frac{\sin x-\sin a-\cos a(x-a)}{% (x-a)^{2}})^{2}-\left( \frac{\cos x-\cos a+\sin a(x-a)}{(x-a)^{2}}\right) +2(% \frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}}{(x-a)^{3}})}{% \left( \frac{\sin x-\sin a}{x-a}\right) ^{2}} \end{eqnarray*} Now using the standard limits

\begin{eqnarray*} \lim_{x\rightarrow a}\frac{\sin x-\sin a}{x-a} &=&\cos a \\ \lim_{x\rightarrow a}\frac{\sin x-\sin a-\left( \cos a\right) (x-a)}{% (x-a)^{2}} &=&-\frac{1}{2}\sin a \\ \lim_{x\rightarrow a}\frac{\cos x-\cos a+\left( \sin a\right) (x-a)}{% (x-a)^{2}} &=&-\frac{1}{2}\cos a \\ \lim_{x\rightarrow a}\frac{\sin x-\sin a-\left( \cos a\right) (x-a)+\frac{1}{% 2}\left( \sin a\right) (x-a)^{2}}{(x-a)^{3}} &=&-\frac{1}{6}\cos a \end{eqnarray*} one gets \begin{eqnarray*} \lim_{x\rightarrow a}f^{\prime }(x) &=&\frac{\sec a(-\frac{1}{2}\sin a)^{2}-\left( -\frac{1}{2}\cos a\right) +2(-\frac{1}{6}\cos a)}{\left( \cos a\right) ^{2}} \\ &=&\frac{\frac{1}{4}\sec a\sin ^{2}a+\frac{1}{2}\cos a-\frac{1}{3}\cos a}{% \cos ^{2}a} \\ &=&\frac{\frac{1}{4}\sec a(1-\cos ^{2}a)+\frac{1}{6}\cos a}{\cos ^{2}a} \\ &=&\frac{\frac{1}{4}\sec a-\frac{1}{4}\cos a+\frac{1}{6}\cos a}{\cos ^{2}a} \\ &=&\frac{\frac{1}{4}\sec a-\frac{1}{12}\cos a}{\cos ^{2}a} \\ &=&\frac{\sec ^{3}a}{4}-\frac{\sec a}{12}. As\ was\ to\ be\ shown! \end{eqnarray*}

$\color{red}{\bf{UPDATE}}$

Here is a proof that \begin{equation*} \lim_{x\rightarrow a}\frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}}{(x-a)^{3}}=-\frac{1}{6}(\cos a) \end{equation*} which does not use L'HOSPITAL's rule, and uses the following facts: \begin{equation*} \begin{array}{ll} \text{Fact 1. } & -1\left. \leq \right. \sin x\left. \leq \right. 1,\ \ \ x\in %TCIMACRO{\U{211d} } %BeginExpansion \mathbb{R} %EndExpansion \\ \text{Fact 2.} & f(t)\leq g(t),\ t\in \left[ a,b\right] \Longrightarrow \int_{a}^{b}f(t)dt\leq \int_{a}^{b}g(t)dt \\ \text{Fact 3.} & \text{Fondamental\ theorem of calculus:}\ \int_{a}^{b}F^{\prime }(t)dt=F(b)-F(a). \\ \text{Fact 4.} & \left( \sin x\right) ^{\prime }=\cos x\ \ \ \ \ and\ \ \ \ \ \left( \cos x\right) ^{\prime }=-\sin x \\ \text{Fact 5.} & \text{Squeeze theorem.} \end{array} \end{equation*} Actually, I assume that $x>a$ and prove that \begin{equation*} \lim_{x\rightarrow a^{+}}\frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}}{(x-a)^{3}}=-\frac{1}{6}(\cos a). \end{equation*} Some easy adaptations are in order to handle the case $x<a$. \begin{eqnarray*} -1\left. \leq \right. \sin x\left. \leq \right. 1 &\Longrightarrow &\int_{a}^{x}(-1)dt\leq \int_{a}^{x}\sin tdt\leq \int_{a}^{x}dt \\ &\Longrightarrow &-(x-a)\leq -\cos x+\cos a\leq (x-a) \\ &\Longrightarrow &-\int_{a}^{x}(t-a)dt\leq \int_{a}^{x}(-\cos t+\cos a)dt\leq \int_{a}^{x}(t-a)dt \\ &\Longrightarrow &-\frac{1}{2}(x-a)^{2}\leq -\sin x+\sin a+\cos a(x-a)\leq \frac{1}{2}(x-a)^{2} \\ &\Longrightarrow &-\frac{1}{2}\int_{a}^{x}(t-a)^{2}dt \\ &\leq &\int_{a}^{x}(-\sin t+\sin a+(\cos a)(t-a))dt \\ &\leq &\frac{1}{2}\int_{a}^{x}(t-a)^{2}dt \\ &\Longrightarrow &-\frac{1}{6}(x-a)^{3}\leq \cos x-\cos a+\sin a(x-a)+\frac{1% }{2}(\cos a)(x-a)^{2} \\ &\leq &\frac{1}{6}(x-a)^{3} \\ &\Longrightarrow &-\frac{1}{6}\int_{a}^{x}(t-a)^{3}dt\leq \int_{a}^{x}(\cos t-\cos a+\sin a(t-a)+\frac{1}{2}(\cos a)(t-a)^{2})dt \\ &\leq &\frac{1}{6}\int_{a}^{x}(t-a)^{3} \\ &\Longrightarrow &-\frac{1}{24}(x-a)^{4} \\ &\leq &\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}+\frac{1}{6}(\cos a)(x-a)^{3} \\ &\leq &\frac{1}{24}(x-a)^{4} \\ &\Longrightarrow &-\frac{1}{24}(x-a)^{4}-\frac{1}{6}(\cos a)(x-a)^{3} \\ &\leq &\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2} \\ &\leq &-\frac{1}{6}(\cos a)(x-a)^{3}+\frac{1}{24}(x-a)^{4} \\ &\Longrightarrow &-\frac{1}{24}(x-a)^{3}-\frac{1}{6}(\cos a) \\ &\leq &\frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}}{(x-a)^{3}} \\ &\leq &-\frac{1}{6}(\cos a)+\frac{1}{24}(x-a) \\ &\Longrightarrow &\lim_{x\rightarrow a^{+}}\left( -\frac{1}{24}(x-a)^{3}-% \frac{1}{6}(\cos a)\right) \\ &\leq &\lim_{x\rightarrow a^{+}}\frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}% \sin a(x-a)^{2}}{(x-a)^{3}} \\ &\leq &\lim_{x\rightarrow a^{+}}-\frac{1}{6}(\cos a)+\frac{1}{24}(x-a) \end{eqnarray*} By squeeze theorem it follows that \begin{equation*} \lim_{x\rightarrow a^{+}}\frac{\sin x-\sin a-\cos a(x-a)+\frac{1}{2}\sin a(x-a)^{2}}{(x-a)^{3}}=-\frac{1}{6}(\cos a). \end{equation*}

@Paramanand, This computations prove at once the existence of the limit and provide its value. It do not use LHR but I agree with you that it make use of matters (Facts 2 and 3) [discussed later in Hardy's book].

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  • $\begingroup$ Fully agree with this approach. Will try to work out P2, P3 $\endgroup$ – Paramanand Singh May 27 '15 at 6:17
  • $\begingroup$ Yes, but P2 is solved yet! I think that P3 can be solved that same way. Let us try it in a moment! $\endgroup$ – Idris May 27 '15 at 6:19
  • $\begingroup$ @Paramanand. The original problem is then completely solved. (No use of Taylor series nor L'hospital's rule as you asked). $\endgroup$ – Idris May 27 '15 at 7:23
  • $\begingroup$ the standard limits whose values are $(-1/2)\sin a, (-1/2)\cos a, (-1/6)\cos a$ do require the use of LHR or Taylor I guess. $\endgroup$ – Paramanand Singh May 27 '15 at 8:28
  • $\begingroup$ On a further note I think we get the limits $(-1/2)\sin a, (-1/2)\cos a$ without LHR or Taylor, but the one with $(-1/6)\cos a$ seems to require Taylor or LHR $\endgroup$ – Paramanand Singh May 27 '15 at 8:33
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$$\begin{align} \lim_{x \to a} f(x) &= \lim_{x\to a}\frac{1}{\sin x - \sin a} - \frac{1}{(x - a)\cos a} \\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)}\\= \ & \sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)} \end{align}$$

Let $ y = x - a$; Since $\sin x - \sin a = 2\sin \left(\dfrac{x-a}{2}\right) \cos \left(\dfrac{x+a}{2}\right)$

$$\begin{align}\sec a\lim_{x\to a}\dfrac{(x-a)\cos a + \sin a - \sin x}{(x-a) (\sin x - \sin a)} & = \dfrac{\sec^2 a}{2}\lim_{y \to 0} \dfrac{(y - \sin y) \cos a + \sin a (1 - \cos y)}{y \sin(y/2)} \\=\ & \sec^2 a\lim_{y \to 0} \dfrac{(y - \sin y) \cos a}{y^2} + \sec^2 a\lim_{y\to0 }\dfrac{\sin a (1 - \cos y)}{y^2 }. \end{align}$$

It is common knowledge that $\lim_{y \to 0} \dfrac{y -\sin y}{y^2} = 0$ and $\lim_{y\to0 }\dfrac{ 1 - \cos y}{y^2 } = \dfrac12$ can be proved only using standard limits(I omitted the proof for the sake of brevity).

Therefore $$\bbox[5px,border-style: solid; border-color:black; border-width: 2px]{\lim_{x \to a} f(x) = \dfrac12 \sec a \tan a.} \tag 1$$


$$f'(x) = \dfrac{1}{(x-a)\cos a} - \dfrac{\cos x}{(\sin x - \sin a)^2}$$

$$\begin{align}\lim_{x \to a} f'(x) = & \sec a \lim_{x\to a}\dfrac{(\sin x -\sin a)^2 - (x-a)^2\cos x \cos a }{(x-a)^2 (\sin x - \sin a)^2} \\= \ & \dfrac{\sec^3 a}{4} \lim_{x\to a}\dfrac{(\sin x -\sin a)^2 - (x-a)^2\cos x \cos a }{(x-a)^2 \sin^2\left(\dfrac{x-a}{2}\right)} \end{align}$$

Letting $ y = x - a$ the above limit becomes,

$$\begin{align}\sec^3 a\lim_{y \to 0} \dfrac{(\sin(y + a ) - \sin a)^2 - y^2 \cos (y +a )\cos a}{y^4}. \end{align}$$

Expanding the numerator completely in terms of $\sin a, \cos a, \sin y $ and $\cos y$, we get

$$\begin{align}\sec^3 a \lim_{y \to 0} \dfrac{\sin^2 y \cos^2 a + \cos^2 y \sin^2 a + \sin^2 a + 2\sin a \cos a \sin y \cos y \\- 2 \sin a \cos a \sin y - 2\sin^2 a \cos y - y^2 \cos y \cos a^2 + y^2 \sin y \sin a \cos a}{y^4}\end{align}$$

Collecting terms simplifies this to $$\sec^3 a\lim_{y\to 0} \dfrac{\cos^2 a(\sin^2 y - y^2 \cos y) + \sin a \cos a \sin y (2 \cos y - 2+ y^2) + \sin^2 a (\cos y - 1)^2}{y^4}$$

  • Solving $\lim_{y \to 0} \dfrac{\sin^2 y - y^2 \cos y}{y^4}$ :

$$\begin{align} \lim_{y \to 0} \dfrac{\sin^2 y - y^2 \cos y}{y^4} =& \lim_{y \to 0} \dfrac{\sin^2 y - y^2 }{y^4}+\lim_{y \to 0}\dfrac{1- \cos y}{y^2} \\= \ &\lim_{y \to 0} \left(\dfrac{\sin y - y }{y^3}\right)\left(\dfrac{\sin y + y }{y}\right)+\lim_{y \to 0}\dfrac{1- \cos y}{y^2}\end{align}$$

Using $\lim_{y \to 0} \dfrac{\sin y - y}{y^3} = \dfrac{-1}{6}$(proved here using only standard limits) and couple other familiar limits, we get the limit as $\dfrac 16$.

  • Solving $\lim_{y \to 0} \sin y\dfrac{(2 \cos y - 2 + y^2)}{y^4}$:

$$\begin{align}\lim_{y \to 0} \sin y\dfrac{(2 \cos y - 2 + y^2)}{y^4} = &\lim_{y \to 0} \dfrac{\sin y}{y}\left(\dfrac{ y^2 - 4\sin^2(y/2)}{y^3}\right) \\=\ &\lim_{y \to 0} \dfrac14\left(\dfrac{ y/2 - 2\sin(y/2)}{(y/2)^3}\right)(y + 2 \sin(y/2)) \\ = \ & \dfrac16 \cdot 0 = 0. \end{align}$$

  • Lastly we use our "common knowledge" to show to conclude that $\lim_{y \to 0} \dfrac{(\cos y - 1)^2}{y^4} = \dfrac14$.

Using above three bullet points we get $$\lim_{x\to a} f'(x) = \dfrac{\sin^2 a\sec^3 a}{4} + \dfrac{\sec a}{6} = \dfrac{\sec^3 a}{4} - \dfrac{\sec a}{12}.$$

Hence $$\bbox[5px,border-style: solid; border-color:black; border-width: 2px]{\lim_{x \to a} f'(x) = \dfrac{\sec^3 a}{4} - \dfrac{\sec a}{12}.} \tag 2$$


Using $(1)$ and $(2)$

$$\begin{align}\dfrac{d}{ da}\{\lim_{x \to a}f(x)\} - \lim_{x \to a}f'(x)=& \sec^3 a - \dfrac12\sec a - \dfrac{\sec^3 a}{4} + \dfrac{\sec a}{12}\\ = \ & \dfrac{3\sec^3 a}{4} - \dfrac{5 \sec a}{12}.\end{align}$$

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  • $\begingroup$ Very good answer with reasonable computational detail. +1 $\endgroup$ – Paramanand Singh Jan 24 '18 at 15:44
  • $\begingroup$ @ParamanandSingh Thank you very much, I have been trying to solve this question for past 2 or 3 days. I have not read Hardy's text but I think the limits I used are basic enough to meet the criteria of this question. $\endgroup$ – user8277998 Jan 24 '18 at 15:46

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