Calculus Examples

$\frac{lim_{t \to - 3} t^{2} - 9}{2 t^{2} + 7 t + 3}$

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 $t$ approaches $- 3$ .

$\frac{lim_{t \to - 3} t^{2} - lim_{t \to - 3} 9}{2 t^{2} + 7 t + 3}$

Step 1.2

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

$\frac{{(lim_{t \to - 3} t)}^{2} - lim_{t \to - 3} 9}{2 t^{2} + 7 t + 3}$

Step 1.3

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

$\frac{{(lim_{t \to - 3} t)}^{2} - 1 \cdot 9}{2 t^{2} + 7 t + 3}$

Step 2

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

$\frac{{(- 3)}^{2} - 1 \cdot 9}{2 t^{2} + 7 t + 3}$

Step 3

Simplify the answer.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

$\frac{9 - 1 \cdot 9}{2 t^{2} + 7 t + 3}$

Step 3.1.2

Multiply $- 1$ by $9$ .

$\frac{9 - 9}{2 t^{2} + 7 t + 3}$

Step 3.1.3

Subtract $9$ from $9$ .

$\frac{0}{2 t^{2} + 7 t + 3}$

Step 3.2

Factor by grouping.

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Step 3.2.1

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 $a \cdot c = 2 \cdot 3 = 6$ and whose sum is $b = 7$ .

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Step 3.2.1.1

Factor $7$ out of $7 t$ .

$\frac{0}{2 t^{2} + 7 (t) + 3}$

Step 3.2.1.2

Rewrite $7$ as $1$ plus $6$

$\frac{0}{2 t^{2} + (1 + 6) t + 3}$

Step 3.2.1.3

Apply the distributive property.

$\frac{0}{2 t^{2} + 1 t + 6 t + 3}$

Step 3.2.1.4

Multiply $t$ by $1$ .

$\frac{0}{2 t^{2} + t + 6 t + 3}$

Step 3.2.2

Factor out the greatest common factor from each group.

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Step 3.2.2.1

Group the first two terms and the last two terms.

$\frac{0}{(2 t^{2} + t) + 6 t + 3}$

Step 3.2.2.2

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

$\frac{0}{t (2 t + 1) + 3 (2 t + 1)}$

Step 3.2.3

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

$\frac{0}{(2 t + 1) (t + 3)}$

Step 3.3

Divide $0$ by $(2 t + 1) (t + 3)$ .

$0$