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Learn AP and College Calculus Limits

Find limits

Try substituting first.

AP and College Calculus: Limits

Work through it

$lim_{x \to 2} (x^{2} + 1)$

Substitute: $4 + 1 = 5$

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

Substituting gives $\frac{0}{0}$ .

Factor: $\frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$

The limit is 6.

Watch out for this

$\frac{0}{0}$ means the limit is 0.

It means you need to keep working.

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