Calculus Examples

$lim_{x \to \frac{π}{2}} (\cos (π \sin (x))) + \sin (π \cos (x))$

Step 1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $\frac{π}{2}$ .

$lim_{x \to \frac{π}{2}} \cos (π \sin (x)) + lim_{x \to \frac{π}{2}} \sin (π \cos (x))$

Step 2

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

$\cos (lim_{x \to \frac{π}{2}} π \sin (x)) + lim_{x \to \frac{π}{2}} \sin (π \cos (x))$

Step 3

Move the term $π$ outside of the limit because it is constant with respect to $x$ .

$\cos (π lim_{x \to \frac{π}{2}} \sin (x)) + lim_{x \to \frac{π}{2}} \sin (π \cos (x))$

Step 4

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

$\cos (π \sin (lim_{x \to \frac{π}{2}} x)) + lim_{x \to \frac{π}{2}} \sin (π \cos (x))$

Step 5

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

$\cos (π \sin (lim_{x \to \frac{π}{2}} x)) + \sin (lim_{x \to \frac{π}{2}} π \cos (x))$

Step 6

Move the term $π$ outside of the limit because it is constant with respect to $x$ .

$\cos (π \sin (lim_{x \to \frac{π}{2}} x)) + \sin (π lim_{x \to \frac{π}{2}} \cos (x))$

Step 7

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

$\cos (π \sin (lim_{x \to \frac{π}{2}} x)) + \sin (π \cos (lim_{x \to \frac{π}{2}} x))$

Step 8

Evaluate the limits by plugging in $\frac{π}{2}$ for all occurrences of $x$ .

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Step 8.1

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

$\cos (π \sin (\frac{π}{2})) + \sin (π \cos (lim_{x \to \frac{π}{2}} x))$

Step 8.2

Evaluate the limit of $x$ by plugging in $\frac{π}{2}$ for $x$ .

$\cos (π \sin (\frac{π}{2})) + \sin (π \cos (\frac{π}{2}))$

Step 9

Simplify the answer.

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Step 9.1

Simplify each term.

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Step 9.1.1

The exact value of $\sin (\frac{π}{2})$ is $1$ .

$\cos (π \cdot 1) + \sin (π \cos (\frac{π}{2}))$

Step 9.1.2

Multiply $π$ by $1$ .

$\cos (π) + \sin (π \cos (\frac{π}{2}))$

Step 9.1.3

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant. Make the expression negative because cosine is negative in the second quadrant.

$- \cos (0) + \sin (π \cos (\frac{π}{2}))$

Step 9.1.4

The exact value of $\cos (0)$ is $1$ .

$- 1 \cdot 1 + \sin (π \cos (\frac{π}{2}))$

Step 9.1.5

Multiply $- 1$ by $1$ .

$- 1 + \sin (π \cos (\frac{π}{2}))$

Step 9.1.6

The exact value of $\cos (\frac{π}{2})$ is $0$ .

$- 1 + \sin (π \cdot 0)$

Step 9.1.7

Multiply $π$ by $0$ .

$- 1 + \sin (0)$

Step 9.1.8

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

$- 1 + 0$

Step 9.2

Add $- 1$ and $0$ .

$- 1$