Calculus Examples

$\frac{lim_{x \to 8} x + 3}{\sqrt{16 x^{2} + 8 x + 1}}$

Step 1

Evaluate the limit.

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Step 1.1

Split the limit using the Sum of Limits Rule on the limit as $x$ approaches $8$ .

$\frac{lim_{x \to 8} x + lim_{x \to 8} 3}{\sqrt{16 x^{2} + 8 x + 1}}$

Step 1.2

Evaluate the limit of $3$ which is constant as $x$ approaches $8$ .

$\frac{lim_{x \to 8} x + 3}{\sqrt{16 x^{2} + 8 x + 1}}$

Step 2

Evaluate the limit of $x$ by plugging in $8$ for $x$ .

$\frac{8 + 3}{\sqrt{16 x^{2} + 8 x + 1}}$

Step 3

Simplify the answer.

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Step 3.1

Add $8$ and $3$ .

$\frac{11}{\sqrt{16 x^{2} + 8 x + 1}}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Factor using the perfect square rule.

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Step 3.2.1.1

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$\frac{11}{\sqrt{{(4 x)}^{2} + 8 x + 1}}$

Step 3.2.1.2

Rewrite $1$ as $1^{2}$ .

$\frac{11}{\sqrt{{(4 x)}^{2} + 8 x + 1^{2}}}$

Step 3.2.1.3

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

$8 x = 2 \cdot (4 x) \cdot 1$

Step 3.2.1.4

Rewrite the polynomial.

$\frac{11}{\sqrt{{(4 x)}^{2} + 2 \cdot (4 x) \cdot 1 + 1^{2}}}$

Step 3.2.1.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 4 x$ and $b = 1$ .

$\frac{11}{\sqrt{{(4 x + 1)}^{2}}}$

Step 3.2.2

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

$\frac{11}{4 x + 1}$