Calculus Examples

$\frac{lim_{n \to 8} 5 n^{2}}{e^{2 n} - 2 n + 10}$

Step 1

Evaluate the limit.

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Step 1.1

Move the term $5$ outside of the limit because it is constant with respect to $n$ .

$\frac{5 lim_{n \to 8} n^{2}}{e^{2 n} - 2 n + 10}$

Step 1.2

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

$\frac{5 {(lim_{n \to 8} n)}^{2}}{e^{2 n} - 2 n + 10}$

Step 2

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

$\frac{5 \cdot 8^{2}}{e^{2 n} - 2 n + 10}$

Step 3

Simplify the answer.

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Step 3.1

Raise $8$ to the power of $2$ .

$\frac{5 \cdot 64}{e^{2 n} - 2 n + 10}$

Step 3.2

Reorder terms.

$\frac{5 \cdot 64}{- 2 n + e^{2 n} + 10}$

Step 3.3

Multiply $5$ by $64$ .

$\frac{320}{- 2 n + e^{2 n} + 10}$

Step 3.4

Factor $- 1$ out of $- 2 n$ .

$\frac{320}{- (2 n) + e^{2 n} + 10}$

Step 3.5

Factor $- 1$ out of $e^{2 n}$ .

$\frac{320}{- (2 n) - 1 (- e^{2 n}) + 10}$

Step 3.6

Factor $- 1$ out of $- (2 n) - 1 (- e^{2 n})$ .

$\frac{320}{- (2 n - e^{2 n}) + 10}$

Step 3.7

Rewrite $10$ as $- 1 (- 10)$ .

$\frac{320}{- (2 n - e^{2 n}) - 1 (- 10)}$

Step 3.8

Factor $- 1$ out of $- (2 n - e^{2 n}) - 1 (- 10)$ .

$\frac{320}{- (2 n - e^{2 n} - 10)}$

Step 3.9

Rewrite $- (2 n - e^{2 n} - 10)$ as $- 1 (2 n - e^{2 n} - 10)$ .

$\frac{320}{- 1 (2 n - e^{2 n} - 10)}$

Step 3.10

Move the negative in front of the fraction.

$- \frac{320}{2 n - e^{2 n} - 10}$