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What is the limit? lim_(xrarr2) (cos(pi/x))/(x-2)

Feb 19, 2018

${\lim}_{x \to 2} \cos \frac{\frac{\pi}{x}}{x - 2} = \frac{\pi}{4}$

Explanation:

Since we get $\frac{0}{0}$ after plugging $2$ in the place of $x$, we can us the L'Hospital's Rule.

L'Hospital's Rule: ${\lim}_{x \to c} \frac{f \left(x\right)}{g \left(x\right)} = \frac{f ' \left(c\right)}{g ' \left(c\right)}$ when you get $f \left(c\right)$ gives you an indeterminate form.

$\implies \frac{\frac{d}{\mathrm{dx}} \left(\cos \left(\frac{\pi}{x}\right)\right)}{\frac{d}{\mathrm{dx}} \left(x - 2\right)}$

I am assuming you know the trigonometric relationships, the power rule, and the chain rule.

$\implies \frac{- \sin \left(\frac{\pi}{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(\pi {x}^{- 1}\right)}{1}$

$\implies - \sin \left(\frac{\pi}{x}\right) \cdot - \pi {x}^{- 2}$

$\implies \sin \left(\frac{\pi}{x}\right) \pi {x}^{-} 2$

We now plug 2 in the place of $x$.

$\implies \sin \left(\frac{\pi}{2}\right) \pi \cdot {2}^{-} 2$

$\implies 1 \cdot \pi \cdot \frac{1}{4}$

$\implies \frac{\pi}{4}$

Feb 19, 2018

${\lim}_{x \rightarrow 2} \frac{\cos \left(\frac{\pi}{x}\right)}{x - 2} = \frac{\pi}{4}$

Explanation:

We seek:

$L = {\lim}_{x \rightarrow 2} \frac{\cos \left(\frac{\pi}{x}\right)}{x - 2}$

Method 1:

Both the numerator or denominator are zero at $x = 2$, and so we have an indeterminate form $\frac{0}{0}$and we can apply L'Hôpital's rule, which states that if we have an indeterminate form, and the limit exists, then:

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

And so:

$L = {\lim}_{x \rightarrow 2} \frac{\frac{d}{\mathrm{dx}} \cos \left(\frac{\pi}{x}\right)}{\frac{d}{\mathrm{dx}} x - 2}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow 2} \frac{- \sin \left(\frac{\pi}{x}\right) \cdot \left(- \frac{\pi}{x} ^ 2\right)}{1}$
$\setminus \setminus \setminus = {\lim}_{x \rightarrow 2} \left(\pi \sin \frac{\frac{\pi}{x}}{x} ^ 2\right)$
$\setminus \setminus \setminus = \pi \sin \frac{\frac{\pi}{2}}{2} ^ 2$
$\setminus \setminus \setminus = \frac{\pi}{4}$

Feb 19, 2018

Without L'Hospital's rule see below.

Explanation:

$\cos A = \sin \left(\frac{\pi}{2} - A\right)$ so we have

$\cos \frac{\frac{\pi}{x}}{x - 2} = \sin \frac{\frac{\pi}{2} - \frac{\pi}{x}}{x - 2}$

$= \sin \frac{\frac{\pi \left(x - 2\right)}{2 x}}{x - 2}$

 = pi/(2x) (sin((pi(x-2))/(2x)))/((pi(x-2))/(2x)

As $x \rightarrow 2$, we have $\left(\frac{\pi \left(x - 2\right)}{2 x}\right) \rightarrow 0$ so we get

${\lim}_{x \rightarrow 2} \frac{\pi}{2 x} \frac{\sin \left(\frac{\pi \left(x - 2\right)}{2 x}\right)}{\frac{\pi \left(x - 2\right)}{2 x}} = \frac{\pi}{2 \left(2\right)} \left(1\right) = \frac{\pi}{4}$