EXAMPLE 4 Evaluate the limit below. lim h → 0 (5 + h)2 − 25h SOLUTION If we define F(h) = (5 + h)2 − 25h then we can't compute the limit by letting h = 0 since F(0) is undefined. But if we simplify F(h) algebraically and simplify at each step, we find that F(h) = + 10h + 25 − 25h
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Step 1: Start with the expression for F(h): \[ F(h) = (5 + h)^2 - 25h \] Show more…
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EXAMPLE 5 Evaluate the limit below: lim (5 + h)^2 - 25 SOLUTION: If we define F(h) = (5 + h)^2 - 25, then we can compute the limit by letting h approach 0. However, F(0) is undefined. But if we simplify F(h) algebraically and simplify at each step, we find that F(h) = (5 + h)^2 - 25 = 25 + 10h + h^2 - 25 = 10h + h^2 (Recall that we consider only h when letting it approach 0). Thus, the limit is...
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Evaluate the limit below. lim h→0 SOLUTION If we define F(h), then we can't compute the limit of F(h) by letting h = 0, because F(0) is undefined. But if we simplify F(h) algebraically, we find the following: F(h) = 4h (Recall that we consider only h approaching 0.) Thus, we get the following limit: lim h→0 4h
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Use algebra to evaluate the limit: lim h->0 ((-5+h)^2 - 25) / h Enter the exact answer: lim h->0 ((-5+h)^2 - 25) / h =
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