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This limit is one of the "Problems Plus" from Stewart Calculus:

$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$

Note that the limit is of the indeterminate form $\frac{0}{0}$. The problem appears several chapters before L'Hôpital's rule is discussed, so I assume there is a solution without using L'Hopital.

Looking at a graph, the local behavior of the function near $0$ appears to be $-\sin(a+x)$, which, of course, suggests a limit of $-\sin(a)$. Using L'Hôpital's rule twice confirms this guess:

$$\begin{align} & \lim_{x \to 0} \frac{\frac{d}{dx} [\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)]}{ \frac{d}{dx} x^{2}} \\ & = \lim_{x \to 0} \frac{2\cos(a+2x)-2\cos(a+x)}{2x} \\ & = \lim_{x \to 0} \frac{-4\sin(a+2x)+2\sin(a+x)}{2} \\ & = -\sin(a). \end{align}$$

Can anyone give a hint or solution for evaluating this limit without L'Hôpital?

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    $\begingroup$ Are you sure about that $\sin(a+x)-2\sin(a+x)$? $\endgroup$ Commented Apr 25, 2020 at 17:56
  • $\begingroup$ @JoséCarlosSantos whoops, the first terms is $\sin(a+2x)$. I will modify my question, but my whole discussion is invalid now. Should I throw it out? $\endgroup$
    – Math2718
    Commented Apr 25, 2020 at 17:58
  • $\begingroup$ Just edit your question. $\endgroup$ Commented Apr 25, 2020 at 18:04

2 Answers 2

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Let's first expand the $\sin$ functions using the addition formulae, $$\sin(a+2x) = \sin(a) \cos(2x) + \cos(a) \sin(2x)$$ and similarly $$\sin(a+x) = \sin a \cos x + \cos a \sin x$$ Expand $x$-dependent $\sin$ and $\cos$ terms to second order, $$\sin x = x + O(x^3), \quad \cos x = 1 - \frac{x^2}{2} + O(x^4)$$ so that we have (writing $S = \sin a$ and $C = \cos a$ for readability) $$\sin(a+2x) = S \cdot \left(1-2x^2\right) + 2Cx + o(x^2)$$ and $$\sin(a+x) = S \cdot \left(1-\frac{x^2}{2}\right) + Cx + o(x^2)$$ Now substitute these into the numerator of interest,

\begin{align*} \sin(a+2x)-2\sin(a+x)+\sin(a) &= S \cdot\left(1-2x^2\right) + 2Cx - 2\left[S\left(1-\frac{x^2}{2}\right)+Cx\right]+S+o(x^2) \\ &= \left(S - 2S + S\right) + x\left[ 2C-2C \right] + x^2 \left[ -2S + S \right] + o(x^2) \\ &= -Sx^2 + o(x^2) \end{align*} where we've collected coefficients by powers of $x$ (which is a somewhat messy experience, but it collects nicely) and hence we have $$\lim_{x \to 0} \frac{\sin(a+2x)-2\sin(a+x)+\sin(a)}{x^2} = \lim_{x \to 0} \left[ -S + \frac{o(x^2)}{x^2} \right] = -S$$ which gives the desired result.


As a side note, L'Hôpital is essentially doing all this expansion and cancelling for us by the machinery of differentiation. In particular, if $f(x) = Ax^2 + o(x^2)$, then $$\frac{f(x)}{x^2} \to A \text{ as } x \to 0$$ which we can "extract" by noting that $$f''(0) = 2A$$ so that $$\lim_{x \to 0} \frac{\frac{d^2}{dx^2} f(x)}{\frac{d^2}{dx^2} x^2} = \lim_{x \to 0} \frac{2A + o(1)}{2} = A$$ Of course, the true story is somewhat more involved (since denominators need not be $x^k$), but I find this a useful intuition to have. If you know about complex residues, I also like to think of this as essentially analogous to residues at higher order poles (since we're extracting a coefficient of an expansion).

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Without heavy machinery, only that $\lim_{x \to 0} \sin x / x = 1$, you can use the sum-to-product formulas, as follows. \begin{align*} \sin(a+2x) - 2 \sin(a+x) + \sin a & = \sin(a+2x) - \sin(a+x) + \sin a - \sin(a+x) \\ & = 2 \sin(x/2) \cos(a + 3x/2) - 2 \sin (x/2) \cos(a + x/2) \\ & = 2 \sin(x/2) \left (\cos(a + 3x/2) - \cos(a + x/2) \right ) \\ & = - 4 \sin(x/2) \sin(a+x) \sin(x/2) \\ & = - 4 \sin^2(x/2) \sin(a+x). \end{align*}

Then $$ \lim_{x \to 0} \frac{\sin^2(x/2)}{x^2} = \frac14 \lim_{x \to 0} \frac{\sin^2(x/2)}{(x/2)^2} = \frac14 \lim_{y \to 0} \left ( \frac{\sin y}{y} \right )^2 = \frac14, $$ and finally the limit of the expression you are looking at is indeed $$ -4 \times \frac14 \times \sin a = - \sin a. $$

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