Calculus Examples

$\frac{lim_{x \to \frac{π}{3}} \sin (x - \frac{π}{3})}{1 - \cos (2 x)}$

Step 1

Evaluate the limit.

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Step 1.1

Move the limit inside the trig function because sine is continuous.

$\frac{\sin (lim_{x \to \frac{π}{3}} x - \frac{π}{3})}{1 - \cos (2 x)}$

Step 1.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{π}{3}$ .

$\frac{\sin (lim_{x \to \frac{π}{3}} x - lim_{x \to \frac{π}{3}} \frac{π}{3})}{1 - \cos (2 x)}$

Step 1.3

Evaluate the limit of $\frac{π}{3}$ which is constant as $x$ approaches $\frac{π}{3}$ .

$\frac{\sin (lim_{x \to \frac{π}{3}} x - \frac{π}{3})}{1 - \cos (2 x)}$

Step 2

Evaluate the limit of $x$ by plugging in $\frac{π}{3}$ for $x$ .

$\frac{\sin (\frac{π}{3} - \frac{π}{3})}{1 - \cos (2 x)}$

Step 3

Simplify the answer.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Combine the numerators over the common denominator.

$\frac{\sin (\frac{π - π}{3})}{1 - \cos (2 x)}$

Step 3.1.2

Subtract $π$ from $π$ .

$\frac{\sin (\frac{0}{3})}{1 - \cos (2 x)}$

Step 3.1.3

Divide $0$ by $3$ .

$\frac{\sin (0)}{1 - \cos (2 x)}$

Step 3.1.4

The exact value of $\sin (0)$ is $0$ .

$\frac{0}{1 - \cos (2 x)}$

Step 3.2

Divide $0$ by $1 - \cos (2 x)$ .

$0$