Calculus Examples

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

Step 1

Apply L'Hospital's rule.

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Step 1.1

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

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Step 1.1.1

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

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

Step 1.1.2

Evaluate the limit of the numerator.

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Step 1.1.2.1

Evaluate the limit.

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Step 1.1.2.1.1

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

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

Step 1.1.2.1.2

Move the term $3$ outside of the limit because it is constant with respect to $x$ .

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

Step 1.1.2.2

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

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

Step 1.1.2.3

Simplify the answer.

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Step 1.1.2.3.1

Combine $3$ and $\frac{π}{2}$ .

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

Step 1.1.2.3.2

Apply the reference angle by finding the angle with equivalent trig values in the first quadrant.

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

Step 1.1.2.3.3

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

$\frac{0}{lim_{x \to \frac{π}{2}} 1 + \cos (2 x)}$

Step 1.1.3

Evaluate the limit of the denominator.

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Step 1.1.3.1

Evaluate the limit.

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Step 1.1.3.1.1

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

$\frac{0}{lim_{x \to \frac{π}{2}} 1 + lim_{x \to \frac{π}{2}} \cos (2 x)}$

Step 1.1.3.1.2

Evaluate the limit of $1$ which is constant as $x$ approaches $\frac{π}{2}$ .

$\frac{0}{1 + lim_{x \to \frac{π}{2}} \cos (2 x)}$

Step 1.1.3.1.3

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

$\frac{0}{1 + \cos (lim_{x \to \frac{π}{2}} 2 x)}$

Step 1.1.3.1.4

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

$\frac{0}{1 + \cos (2 lim_{x \to \frac{π}{2}} x)}$

Step 1.1.3.2

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

$\frac{0}{1 + \cos (2 \frac{π}{2})}$

Step 1.1.3.3

Simplify the answer.

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Step 1.1.3.3.1

Simplify each term.

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Step 1.1.3.3.1.1

Cancel the common factor of $2$ .

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Step 1.1.3.3.1.1.1

Cancel the common factor.

$\frac{0}{1 + \cos (2 \frac{π}{2})}$

Step 1.1.3.3.1.1.2

Rewrite the expression.

$\frac{0}{1 + \cos (π)}$

Step 1.1.3.3.1.2

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.

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

Step 1.1.3.3.1.3

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

$\frac{0}{1 - 1 \cdot 1}$

Step 1.1.3.3.1.4

Multiply $- 1$ by $1$ .

$\frac{0}{1 - 1}$

Step 1.1.3.3.2

Subtract $1$ from $1$ .

$\frac{0}{0}$

Step 1.1.3.3.3

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.

$lim_{x \to \frac{π}{2}} \frac{\cos (3 x)}{1 + \cos (2 x)} = lim_{x \to \frac{π}{2}} \frac{\frac{d}{d x} [\cos (3 x)]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 1.3

Find the derivative of the numerator and denominator.

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Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to \frac{π}{2}} \frac{\frac{d}{d x} [\cos (3 x)]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 1.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \cos (x)$ and $g (x) = 3 x$ .

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Step 1.3.2.1

To apply the Chain Rule, set $u$ as $3 x$ .

$lim_{x \to \frac{π}{2}} \frac{\frac{d}{d u} [\cos (u)] \frac{d}{d x} [3 x]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 1.3.2.2

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

$lim_{x \to \frac{π}{2}} \frac{- \sin (u) \frac{d}{d x} [3 x]}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 1.3.2.3

Replace all occurrences of $u$ with $3 x$ .

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

Step 1.3.3

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

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

Step 1.3.4

Multiply $3$ by $- 1$ .

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

Step 1.3.5

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x) \cdot 1}{\frac{d}{d x} [1 + \cos (2 x)]}$

Step 1.3.6

Multiply $- 3$ by $1$ .

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

Step 1.3.7

By the Sum Rule, the derivative of $1 + \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [\cos (2 x)]$ .

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

Step 1.3.8

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 + \frac{d}{d x} [\cos (2 x)]}$

Step 1.3.9

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

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Step 1.3.9.1

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

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Step 1.3.9.1.1

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 + \frac{d}{d u} [\cos (u)] \frac{d}{d x} [2 x]}$

Step 1.3.9.1.2

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - \sin (u) \frac{d}{d x} [2 x]}$

Step 1.3.9.1.3

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - \sin (2 x) \frac{d}{d x} [2 x]}$

Step 1.3.9.2

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - \sin (2 x) (2 \frac{d}{d x} [x])}$

Step 1.3.9.3

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

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - \sin (2 x) (2 \cdot 1)}$

Step 1.3.9.4

Multiply $2$ by $1$ .

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - \sin (2 x) \cdot 2}$

Step 1.3.9.5

Multiply $2$ by $- 1$ .

$lim_{x \to \frac{π}{2}} \frac{- 3 \sin (3 x)}{0 - 2 \sin (2 x)}$

Step 1.3.10

Subtract $2 \sin (2 x)$ from $0$ .

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

Step 2

Since the function approaches $- \infty$ from the left and $\infty$ from the right, the limit does not exist.

$Does not exist$