Calculus Examples

$lim_{x \to - 3} \frac{x^{2} + 3 x}{\sqrt{x^{2} + 6 x + 9}}$

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.

$\frac{lim_{x \to - 3} x^{2} + 3 x}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

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 $- 3$ .

$\frac{lim_{x \to - 3} x^{2} + lim_{x \to - 3} 3 x}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.2

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

$\frac{{(lim_{x \to - 3} x)}^{2} + lim_{x \to - 3} 3 x}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.3

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

$\frac{{(lim_{x \to - 3} x)}^{2} + 3 lim_{x \to - 3} x}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.4

Evaluate the limits by plugging in $- 3$ for all occurrences of $x$ .

Tap for more steps...

Step 1.1.2.4.1

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{{(- 3)}^{2} + 3 lim_{x \to - 3} x}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.4.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{{(- 3)}^{2} + 3 \cdot - 3}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.5

Simplify the answer.

Tap for more steps...

Step 1.1.2.5.1

Simplify each term.

Tap for more steps...

Step 1.1.2.5.1.1

Raise $- 3$ to the power of $2$ .

$\frac{9 + 3 \cdot - 3}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.5.1.2

Multiply $3$ by $- 3$ .

$\frac{9 - 9}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.2.5.2

Subtract $9$ from $9$ .

$\frac{0}{lim_{x \to - 3} \sqrt{x^{2} + 6 x + 9}}$

Step 1.1.3

Evaluate the limit of the denominator.

Tap for more steps...

Step 1.1.3.1

Move the limit under the radical sign.

$\frac{0}{\sqrt{lim_{x \to - 3} x^{2} + 6 x + 9}}$

Step 1.1.3.2

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$\frac{0}{\sqrt{lim_{x \to - 3} x^{2} + lim_{x \to - 3} 6 x + lim_{x \to - 3} 9}}$

Step 1.1.3.3

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

$\frac{0}{\sqrt{{(lim_{x \to - 3} x)}^{2} + lim_{x \to - 3} 6 x + lim_{x \to - 3} 9}}$

Step 1.1.3.4

Move the term $6$ outside of the limit because it is constant with respect to $x$ .

$\frac{0}{\sqrt{{(lim_{x \to - 3} x)}^{2} + 6 lim_{x \to - 3} x + lim_{x \to - 3} 9}}$

Step 1.1.3.5

Evaluate the limit of $9$ which is constant as $x$ approaches $- 3$ .

$\frac{0}{\sqrt{{(lim_{x \to - 3} x)}^{2} + 6 lim_{x \to - 3} x + 9}}$

Step 1.1.3.6

Evaluate the limits by plugging in $- 3$ for all occurrences of $x$ .

Tap for more steps...

Step 1.1.3.6.1

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{0}{\sqrt{{(- 3)}^{2} + 6 lim_{x \to - 3} x + 9}}$

Step 1.1.3.6.2

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$\frac{0}{\sqrt{{(- 3)}^{2} + 6 \cdot - 3 + 9}}$

Step 1.1.3.7

Simplify the answer.

Tap for more steps...

Step 1.1.3.7.1

Raise $- 3$ to the power of $2$ .

$\frac{0}{\sqrt{9 + 6 \cdot - 3 + 9}}$

Step 1.1.3.7.2

Multiply $6$ by $- 3$ .

$\frac{0}{\sqrt{9 - 18 + 9}}$

Step 1.1.3.7.3

Subtract $18$ from $9$ .

$\frac{0}{\sqrt{- 9 + 9}}$

Step 1.1.3.7.4

Add $- 9$ and $9$ .

$\frac{0}{\sqrt{0}}$

Step 1.1.3.7.5

Rewrite $0$ as $0^{2}$ .

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

Step 1.1.3.7.6

Pull terms out from under the radical, assuming positive real numbers.

$\frac{0}{0}$

Step 1.1.3.7.7

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

Undefined

$\frac{0}{0}$

Step 1.1.3.8

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.

$lim_{x \to - 3} \frac{x^{2} + 3 x}{\sqrt{x^{2} + 6 x + 9}} = lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} + 3 x]}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3

Find the derivative of the numerator and denominator.

Tap for more steps...

Step 1.3.1

Differentiate the numerator and denominator.

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2} + 3 x]}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3.2

By the Sum Rule, the derivative of $x^{2} + 3 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x]$ .

$lim_{x \to - 3} \frac{\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3 x]}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3.3

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

$lim_{x \to - 3} \frac{2 x + \frac{d}{d x} [3 x]}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3.4

Evaluate $\frac{d}{d x} [3 x]$ .

Tap for more steps...

Step 1.3.4.1

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

$lim_{x \to - 3} \frac{2 x + 3 \frac{d}{d x} [x]}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

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 $n = 1$ .

$lim_{x \to - 3} \frac{2 x + 3 \cdot 1}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3.4.3

Multiply $3$ by $1$ .

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]}$

Step 1.3.5

Rewrite $\frac{d}{d x} [\sqrt{x^{2} + 6 x + 9}]$ as $\frac{d}{d x} [x + 3]$ .

Tap for more steps...

Step 1.3.5.1

Factor using the perfect square rule.

Tap for more steps...

Step 1.3.5.1.1

Rewrite $9$ as $3^{2}$ .

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{x^{2} + 6 x + 3^{2}}]}$

Step 1.3.5.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Step 1.3.5.1.3

Rewrite the polynomial.

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{x^{2} + 2 \cdot x \cdot 3 + 3^{2}}]}$

Step 1.3.5.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [\sqrt{{(x + 3)}^{2}}]}$

Step 1.3.5.2

Pull terms out from under the radical, assuming positive real numbers.

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [x + 3]}$

Step 1.3.6

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

$lim_{x \to - 3} \frac{2 x + 3}{\frac{d}{d x} [x] + \frac{d}{d x} [3]}$

Step 1.3.7

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

$lim_{x \to - 3} \frac{2 x + 3}{1 + \frac{d}{d x} [3]}$

Step 1.3.8

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$lim_{x \to - 3} \frac{2 x + 3}{1 + 0}$

Step 1.3.9

Add $1$ and $0$ .

$lim_{x \to - 3} \frac{2 x + 3}{1}$

Step 1.4

Divide $2 x + 3$ by $1$ .

$lim_{x \to - 3} 2 x + 3$

Step 2

Evaluate the limit.

Tap for more steps...

Step 2.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $- 3$ .

$lim_{x \to - 3} 2 x + lim_{x \to - 3} 3$

Step 2.2

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

$2 lim_{x \to - 3} x + lim_{x \to - 3} 3$

Step 2.3

Evaluate the limit of $3$ which is constant as $x$ approaches $- 3$ .

$2 lim_{x \to - 3} x + 3$

Step 3

Evaluate the limit of $x$ by plugging in $- 3$ for $x$ .

$2 \cdot - 3 + 3$

Step 4

Simplify the answer.

Tap for more steps...

Step 4.1

Multiply $2$ by $- 3$ .

$- 6 + 3$

Step 4.2

Add $- 6$ and $3$ .

$- 3$