Calculus Examples

$lim_{x \to 2} \frac{x^{2} - 4 x + 4}{\sin^{2} (π x)}$

Step 1

Apply L'Hospital's rule.

Tap for more steps...

Step 1.1

Evaluate the limit of the numerator and the limit of the denominator.

Tap for more steps...

Step 1.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{x \to 2} x^{2} - 4 x + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 1.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{lim_{x \to 2} x^{2} - lim_{x \to 2} 4 x + lim_{x \to 2} 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.2

Move the exponent $2$ from $x^{2}$ outside the limit using the Limits Power Rule.

$\frac{{(lim_{x \to 2} x)}^{2} - lim_{x \to 2} 4 x + lim_{x \to 2} 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.3

Move the term $4$ outside of the limit because it is constant with respect to $x$ .

$\frac{{(lim_{x \to 2} x)}^{2} - 4 lim_{x \to 2} x + lim_{x \to 2} 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.4

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

$\frac{{(lim_{x \to 2} x)}^{2} - 4 lim_{x \to 2} x + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.5

Evaluate the limits by plugging in $2$ for all occurrences of $x$ .

Tap for more steps...

Step 1.1.2.5.1

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{2^{2} - 4 lim_{x \to 2} x + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.5.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{2^{2} - 4 \cdot 2 + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.6

Simplify the answer.

Tap for more steps...

Step 1.1.2.6.1

Simplify each term.

Tap for more steps...

Step 1.1.2.6.1.1

Raise $2$ to the power of $2$ .

$\frac{4 - 4 \cdot 2 + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.6.1.2

Multiply $- 4$ by $2$ .

$\frac{4 - 8 + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.6.2

Subtract $8$ from $4$ .

$\frac{- 4 + 4}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.2.6.3

Add $- 4$ and $4$ .

$\frac{0}{lim_{x \to 2} \sin^{2} (π x)}$

Step 1.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.1.3.1

Evaluate the limit.

Tap for more steps...

Step 1.1.3.1.1

Move the exponent $2$ from $\sin^{2} (π x)$ outside the limit using the Limits Power Rule.

$\frac{0}{{(lim_{x \to 2} \sin (π x))}^{2}}$

Step 1.1.3.1.2

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

$\frac{0}{\sin^{2} (lim_{x \to 2} π x)}$

Step 1.1.3.1.3

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

$\frac{0}{\sin^{2} (π lim_{x \to 2} x)}$

Step 1.1.3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{0}{\sin^{2} (π \cdot 2)}$

Step 1.1.3.3

Simplify the answer.

Tap for more steps...

Step 1.1.3.3.1

Move $2$ to the left of $π$ .

$\frac{0}{\sin^{2} (2 \cdot π)}$

Step 1.1.3.3.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{0}{\sin^{2} (0)}$

Step 1.1.3.3.3

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

$\frac{0}{0^{2}}$

Step 1.1.3.3.4

Raising $0$ to any positive power yields $0$ .

$\frac{0}{0}$

Step 1.1.3.3.5

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 1.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 2} \frac{x^{2} - 4 x + 4}{\sin^{2} (π x)} = lim_{x \to 2} \frac{\frac{d}{d x} [x^{2} - 4 x + 4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to 2} \frac{\frac{d}{d x} [x^{2} - 4 x + 4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} - 4 x + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4]$ .

$lim_{x \to 2} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.3

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$lim_{x \to 2} \frac{2 x + \frac{d}{d x} [- 4 x] + \frac{d}{d x} [4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [- 4 x]$ .

Tap for more steps...

Step 1.3.4.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{2 x - 4 \frac{d}{d x} [x] + \frac{d}{d x} [4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.4.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 2} \frac{2 x - 4 \cdot 1 + \frac{d}{d x} [4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.4.3

Multiply $- 4$ by $1$ .

$lim_{x \to 2} \frac{2 x - 4 + \frac{d}{d x} [4]}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.5

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$lim_{x \to 2} \frac{2 x - 4 + 0}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.6

Add $2 x - 4$ and $0$ .

$lim_{x \to 2} \frac{2 x - 4}{\frac{d}{d x} [\sin^{2} (π x)]}$

Step 1.3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = \sin (π x)$ .

Tap for more steps...

Step 1.3.7.1

To apply the Chain Rule, set $u_{1}$ as $\sin (π x)$ .

$lim_{x \to 2} \frac{2 x - 4}{\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [\sin (π x)]}$

Step 1.3.7.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 2$ .

$lim_{x \to 2} \frac{2 x - 4}{2 u_{1} \frac{d}{d x} [\sin (π x)]}$

Step 1.3.7.3

Replace all occurrences of $u_{1}$ with $\sin (π x)$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) \frac{d}{d x} [\sin (π x)]}$

Step 1.3.8

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = π x$ .

Tap for more steps...

Step 1.3.8.1

To apply the Chain Rule, set $u_{2}$ as $π x$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) (\frac{d}{d u_{2}} [\sin (u_{2})] \frac{d}{d x} [π x])}$

Step 1.3.8.2

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) (\cos (u_{2}) \frac{d}{d x} [π x])}$

Step 1.3.8.3

Replace all occurrences of $u_{2}$ with $π x$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) (\cos (π x) \frac{d}{d x} [π x])}$

Step 1.3.9

Since $π$ is constant with respect to $x$ , the derivative of $π x$ with respect to $x$ is $π \frac{d}{d x} [x]$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) \cos (π x) (π \frac{d}{d x} [x])}$

Step 1.3.10

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) \cos (π x) (π \cdot 1)}$

Step 1.3.11

Multiply $π$ by $1$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \sin (π x) \cos (π x) π}$

Step 1.3.12

Simplify.

Tap for more steps...

Step 1.3.12.1

Reorder the factors of $2 \sin (π x) \cos (π x) π$ .

$lim_{x \to 2} \frac{2 x - 4}{2 π \cos (π x) \sin (π x)}$

Step 1.3.12.2

Add parentheses.

$lim_{x \to 2} \frac{2 x - 4}{2 π (\cos (π x) \sin (π x))}$

Step 1.3.12.3

Reorder $2 π$ and $\cos (π x) \sin (π x)$ .

$lim_{x \to 2} \frac{2 x - 4}{\cos (π x) \sin (π x) (2 π)}$

Step 1.3.12.4

Add parentheses.

$lim_{x \to 2} \frac{2 x - 4}{\cos (π x) (\sin (π x) \cdot 2) π}$

Step 1.3.12.5

Reorder $\cos (π x)$ and $\sin (π x) \cdot 2$ .

$lim_{x \to 2} \frac{2 x - 4}{\sin (π x) \cdot 2 \cos (π x) π}$

Step 1.3.12.6

Reorder $\sin (π x)$ and $2$ .

$lim_{x \to 2} \frac{2 x - 4}{2 \cdot \sin (π x) \cos (π x) π}$

Step 1.3.12.7

Apply the sine double-angle identity.

$lim_{x \to 2} \frac{2 x - 4}{\sin (2 (π x)) π}$

Step 1.3.12.8

Reorder factors in $\sin (2 π x) π$ .

$lim_{x \to 2} \frac{2 x - 4}{π \sin (2 π x)}$

Step 2

Move the term $\frac{1}{π}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{π} lim_{x \to 2} \frac{2 x - 4}{\sin (2 π x)}$

Step 3

Apply L'Hospital's rule.

Tap for more steps...

Step 3.1

Evaluate the limit of the numerator and the limit of the denominator.

Tap for more steps...

Step 3.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{1}{π} \cdot \frac{lim_{x \to 2} 2 x - 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2

Evaluate the limit of the numerator.

Tap for more steps...

Step 3.1.2.1

Evaluate the limit.

Tap for more steps...

Step 3.1.2.1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $2$ .

$\frac{1}{π} \cdot \frac{lim_{x \to 2} 2 x - lim_{x \to 2} 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.1.2

Move the term $2$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{π} \cdot \frac{2 lim_{x \to 2} x - lim_{x \to 2} 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.1.3

Evaluate the limit of $4$ which is constant as $x$ approaches $2$ .

$\frac{1}{π} \cdot \frac{2 lim_{x \to 2} x - 1 \cdot 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{1}{π} \cdot \frac{2 \cdot 2 - 1 \cdot 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.3

Simplify the answer.

Tap for more steps...

Step 3.1.2.3.1

Simplify each term.

Tap for more steps...

Step 3.1.2.3.1.1

Multiply $2$ by $2$ .

$\frac{1}{π} \cdot \frac{4 - 1 \cdot 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.3.1.2

Multiply $- 1$ by $4$ .

$\frac{1}{π} \cdot \frac{4 - 4}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.2.3.2

Subtract $4$ from $4$ .

$\frac{1}{π} \cdot \frac{0}{lim_{x \to 2} \sin (2 π x)}$

Step 3.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 3.1.3.1

Evaluate the limit.

Tap for more steps...

Step 3.1.3.1.1

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

$\frac{1}{π} \cdot \frac{0}{\sin (lim_{x \to 2} 2 π x)}$

Step 3.1.3.1.2

Move the term $2 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{π} \cdot \frac{0}{\sin (2 π lim_{x \to 2} x)}$

Step 3.1.3.2

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

$\frac{1}{π} \cdot \frac{0}{\sin (2 π \cdot 2)}$

Step 3.1.3.3

Simplify the answer.

Tap for more steps...

Step 3.1.3.3.1

Multiply $2$ by $2$ .

$\frac{1}{π} \cdot \frac{0}{\sin (4 π)}$

Step 3.1.3.3.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{1}{π} \cdot \frac{0}{\sin (0)}$

Step 3.1.3.3.3

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

$\frac{1}{π} \cdot \frac{0}{0}$

Step 3.1.3.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{π} \cdot \frac{0}{0}$

Step 3.1.3.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{π} \cdot \frac{0}{0}$

Step 3.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{1}{π} \cdot \frac{0}{0}$

Step 3.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{x \to 2} \frac{2 x - 4}{\sin (2 π x)} = lim_{x \to 2} \frac{\frac{d}{d x} [2 x - 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 3.3.1

Differentiate the numerator and denominator.

$\frac{1}{π} lim_{x \to 2} \frac{\frac{d}{d x} [2 x - 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.2

By the Sum Rule, the derivative of $2 x - 4$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4]$ .

$\frac{1}{π} lim_{x \to 2} \frac{\frac{d}{d x} [2 x] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.3

Evaluate $\frac{d}{d x} [2 x]$ .

Tap for more steps...

Step 3.3.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

$\frac{1}{π} lim_{x \to 2} \frac{2 \frac{d}{d x} [x] + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{π} lim_{x \to 2} \frac{2 \cdot 1 + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.3.3

Multiply $2$ by $1$ .

$\frac{1}{π} lim_{x \to 2} \frac{2 + \frac{d}{d x} [- 4]}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$\frac{1}{π} lim_{x \to 2} \frac{2 + 0}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.5

Add $2$ and $0$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\frac{d}{d x} [\sin (2 π x)]}$

Step 3.3.6

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = 2 π x$ .

Tap for more steps...

Step 3.3.6.1

To apply the Chain Rule, set $u$ as $2 π x$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\frac{d}{d u} [\sin (u)] \frac{d}{d x} [2 π x]}$

Step 3.3.6.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\cos (u) \frac{d}{d x} [2 π x]}$

Step 3.3.6.3

Replace all occurrences of $u$ with $2 π x$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\cos (2 π x) \frac{d}{d x} [2 π x]}$

Step 3.3.7

Since $2 π$ is constant with respect to $x$ , the derivative of $2 π x$ with respect to $x$ is $2 π \frac{d}{d x} [x]$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\cos (2 π x) \cdot 2 π \frac{d}{d x} [x]}$

Step 3.3.8

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\cos (2 π x) \cdot 2 π \cdot 1}$

Step 3.3.9

Multiply $2$ by $1$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{\cos (2 π x) \cdot 2 π}$

Step 3.3.10

Move $2$ to the left of $\cos (2 π x)$ .

$\frac{1}{π} lim_{x \to 2} \frac{2}{2 \cdot \cos (2 π x) π}$

Step 3.3.11

Reorder the factors of $2 \cos (2 π x) π$ .

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

Step 3.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.4.1

Cancel the common factor.

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

Step 3.4.2

Rewrite the expression.

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

Step 4

Evaluate the limit.

Tap for more steps...

Step 4.1

Move the term $\frac{1}{π}$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{π} \cdot \frac{1}{π} lim_{x \to 2} \frac{1}{\cos (2 π x)}$

Step 4.2

Split the limit using the Limits Quotient Rule on the limit as $x$ approaches $2$ .

$\frac{1}{π} \cdot \frac{1}{π} \frac{lim_{x \to 2} 1}{lim_{x \to 2} \cos (2 π x)}$

Step 4.3

Evaluate the limit of $1$ which is constant as $x$ approaches $2$ .

$\frac{1}{π} \cdot \frac{1}{π} \frac{1}{lim_{x \to 2} \cos (2 π x)}$

Step 4.4

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

$\frac{1}{π} \cdot \frac{1}{π} \frac{1}{\cos (lim_{x \to 2} 2 π x)}$

Step 4.5

Move the term $2 π$ outside of the limit because it is constant with respect to $x$ .

$\frac{1}{π} \cdot \frac{1}{π} \frac{1}{\cos (2 π lim_{x \to 2} x)}$

Step 5

Evaluate the limit of $x$ by plugging in $2$ for $x$ .

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

Step 6

Simplify the answer.

Tap for more steps...

Step 6.1

Multiply $\frac{1}{π} \cdot \frac{1}{π}$ .

Tap for more steps...

Step 6.1.1

Multiply $\frac{1}{π}$ by $\frac{1}{π}$ .

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

Step 6.1.2

Raise $π$ to the power of $1$ .

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

Step 6.1.3

Raise $π$ to the power of $1$ .

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

Step 6.1.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.1.5

Add $1$ and $1$ .

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

Step 6.2

Combine.

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

Step 6.3

Multiply $1$ by $1$ .

$\frac{1}{π^{2} \cos (2 π \cdot 2)}$

Step 6.4

Simplify the denominator.

Tap for more steps...

Step 6.4.1

Multiply $2$ by $2$ .

$\frac{1}{π^{2} \cos (4 π)}$

Step 6.4.2

Subtract full rotations of $2 π$ until the angle is greater than or equal to $0$ and less than $2 π$ .

$\frac{1}{π^{2} \cos (0)}$

Step 6.4.3

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

$\frac{1}{π^{2} \cdot 1}$

Step 6.5

Multiply $π^{2}$ by $1$ .

$\frac{1}{π^{2}}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$\frac{1}{π^{2}}$

Decimal Form:

$0.10132118 \dots$