Calculus Examples

$\frac{lim_{x \to - 8} t^{2} - 64}{2 t^{2} + 17 t + 8}$

Step 1

Evaluate the limit of $t^{2} - 64$ which is constant as $x$ approaches $- 8$ .

$\frac{t^{2} - 64}{2 t^{2} + 17 t + 8}$

Step 2

Simplify the answer.

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Step 2.1

Simplify the numerator.

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Step 2.1.1

Rewrite $64$ as $8^{2}$ .

$\frac{t^{2} - 8^{2}}{2 t^{2} + 17 t + 8}$

Step 2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 8$ .

$\frac{(t + 8) (t - 8)}{2 t^{2} + 17 t + 8}$

Step 2.2

Factor by grouping.

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Step 2.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot 8 = 16$ and whose sum is $b = 17$ .

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Step 2.2.1.1

Factor $17$ out of $17 t$ .

$\frac{(t + 8) (t - 8)}{2 t^{2} + 17 (t) + 8}$

Step 2.2.1.2

Rewrite $17$ as $1$ plus $16$

$\frac{(t + 8) (t - 8)}{2 t^{2} + (1 + 16) t + 8}$

Step 2.2.1.3

Apply the distributive property.

$\frac{(t + 8) (t - 8)}{2 t^{2} + 1 t + 16 t + 8}$

Step 2.2.1.4

Multiply $t$ by $1$ .

$\frac{(t + 8) (t - 8)}{2 t^{2} + t + 16 t + 8}$

Step 2.2.2

Factor out the greatest common factor from each group.

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Step 2.2.2.1

Group the first two terms and the last two terms.

$\frac{(t + 8) (t - 8)}{(2 t^{2} + t) + 16 t + 8}$

Step 2.2.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{(t + 8) (t - 8)}{t (2 t + 1) + 8 (2 t + 1)}$

Step 2.2.3

Factor the polynomial by factoring out the greatest common factor, $2 t + 1$ .

$\frac{(t + 8) (t - 8)}{(2 t + 1) (t + 8)}$

Step 2.3

Cancel the common factor of $t + 8$ .

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Step 2.3.1

Cancel the common factor.

$\frac{(t + 8) (t - 8)}{(2 t + 1) (t + 8)}$

Step 2.3.2

Rewrite the expression.

$\frac{t - 8}{2 t + 1}$