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I have to find $$\lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$

We are not allowed to use L'Hospital's rule, any suggestions would be beneficial!

I have tried multiplying by the conjugate (both the numerator and the denominator).

I have tried using trig substitution.

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    $\begingroup$ Can't you just recognize this as the derivative of $\sin(x)$ evaluated at $x = \pi/6$? This isn't L'hopital's rule, it's just using the definition of the derivative. $\endgroup$ – J. Marx-Kuo Sep 13 '16 at 0:20
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Let $x'=x-\pi/6$. Then $$\lim_{x' \to 0} \frac{\sin(x'+\pi/6)-\frac{1}{2}}{x'}$$ Then apply the addition formula for sine, and it will reduce to $$\lim_{x' \to 0} \frac{\sin(x')\cos(\pi/6)+\cos(x')sin(\pi/6)-\frac{1}{2}}{x'}$$ $$\lim_{x' \to 0} \frac{\sin(x')\cos(\pi/6)}{x'}+\frac{\cos(x')sin(\pi/6)-\frac{1}{2}}{x'}$$ $$=\frac{\sqrt{3}}{2}+0$$

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HINT: Use the definition for derivatives on the function $f(x) = \sin x$ and use the fact that $\sin(\frac{\pi}{6}) = \frac 12$

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Using the definition of the derivative of $\sin(x)$ at $x = \dfrac{\pi}{6}$, we get

\begin{align} \lim_{x \to \frac{\pi}{6}}\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}} &=\lim_{x \to \frac{\pi}{6}} \frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}} \\ &= \sin'(\frac{\pi}{6}) \\ &= \cos(\frac{\pi}{6}) \\ &= \dfrac{\sqrt 3}{2} \end{align}

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  • $\begingroup$ @IanMiller Thanks. I fixed it. $\endgroup$ – steven gregory Sep 13 '16 at 0:52
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Just another way to do it.

Considering $$A=\frac{\sin(x)-\frac{1}{2}}{x-\frac{\pi}{6}}$$ make life simpler using $x=y+\frac{\pi}{6}$ and expand $\sin(y+\frac{\pi}{6})$. This makes $$A=\frac {\sqrt{3} \sin (y)+\cos (y)-1}{2y}$$ Now, let us Taylor expansions around $y=0$; this gives $$\sqrt{3} \sin (y)+\cos (y)-1=\sqrt{3} y-\frac{y^2}{2}+O\left(y^3\right)$$ and so $$A=\frac{\sqrt{3}}{2}-\frac{y}{4}+O\left(y^2\right)$$ which shows the limit and how it is approached when $y\to 0$.

Edit

If you add the next term of the Taylor expansion, you would get $$A=\frac{\sqrt{3}}{2}-\frac{y}{4}-\frac{y^2}{4 \sqrt{3}}+O\left(y^3\right)$$ You could me amazed to see how close are the original function and the above approximation for the range $-1\leq y \leq 1$. This means that, if, at any time, you need to find $x$ such that $A(x)=a$,$(a>0)$ you would a nice approximation solving for $y$ the quadratic$$a=\frac{\sqrt{3}}{2}-\frac{y}{4}-\frac{y^2}{4 \sqrt{3}}$$ the solution of which being $$y=\frac{ \sqrt{27-16 \sqrt{3} a}-\sqrt 3}{2 }$$ For illustration purposes, let us use $a=\frac 12$; this would give an approximate solution $y\approx 0.9467$ while the "exact" solution would be $y\approx 0.9975$ which is not too bad.

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Hint:

Think of the rate of variation $\;\dfrac{\Delta f}{\Delta x}=\dfrac{f(x')-f(x)}{x'-x}$.

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