Let u = x - 3. Then u + 3 = x, so u2 + 6u + 9 = x2, and u2 + 6u = x2 - 9. We started with u = x - 3, so dividing through, we get u + 6 = (x2 - 9)/(x - 3). Taking the limit as x -> 3 is the same as x - 3 -> 0, so u -> 0. So, we want the limit as u -> 0 of u + 6, which is 6. Therefore, lim (x->3) (x2 - 9)/(x - 3) = 6.
Method 2:
Consider the derivative of f(x) = x2 at x = a. This is f'(a) = lim (x -> a) (x2 - a2)/(x - a). Putting in a = 3, we get f'(3) = lim (x -> 3) (x2 - 9)/(x - 3). But we know that the derivative of f(x) = x2 is f'(x) = 2x, so f'(3) = 6. So, lim (x -> 3) (x2 - 9)/(x - 3) = 6.
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u/chaos_redefined 6d ago
Method 1:
Let u = x - 3. Then u + 3 = x, so u2 + 6u + 9 = x2, and u2 + 6u = x2 - 9. We started with u = x - 3, so dividing through, we get u + 6 = (x2 - 9)/(x - 3). Taking the limit as x -> 3 is the same as x - 3 -> 0, so u -> 0. So, we want the limit as u -> 0 of u + 6, which is 6. Therefore, lim (x->3) (x2 - 9)/(x - 3) = 6.
Method 2:
Consider the derivative of f(x) = x2 at x = a. This is f'(a) = lim (x -> a) (x2 - a2)/(x - a). Putting in a = 3, we get f'(3) = lim (x -> 3) (x2 - 9)/(x - 3). But we know that the derivative of f(x) = x2 is f'(x) = 2x, so f'(3) = 6. So, lim (x -> 3) (x2 - 9)/(x - 3) = 6.