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Secant slope and its limit at x = 1

Consider the function f(x)=2x2−x+1f(x) = 2x^2 - x + 1.

Let f(x)=2x2−x+1f(x) = 2x^2 - x + 1. Consider the secant line through the points (1,f(1))(1, f(1)) and (1+h,f(1+h))(1+h, f(1+h)).

First, write an expression for the slope of this secant line, f(1+h)−f(1)h\dfrac{f(1+h)-f(1)}{h}, simplified as much as possible. Then compute lim⁡h→0f(1+h)−f(1)h\displaystyle\lim_{h\to 0} \dfrac{f(1+h)-f(1)}{h}, and state what this limit represents. What is the value of this limit?

Hint

Expand f(1+h)f(1+h) fully, subtract f(1)f(1), then factor an hh out of the numerator before canceling with the hh in the denominator. This cancellation is what allows the limit as h→0h\to 0 to be evaluated directly.

Step-by-step solution

First compute f(1+h)=2(1+h)2−(1+h)+1=2(1+2h+h2)−1−h+1=2+4h+2h2−h=2h2+3h+2f(1+h) = 2(1+h)^2 - (1+h) + 1 = 2(1+2h+h^2) - 1 - h + 1 = 2 + 4h + 2h^2 - h = 2h^2 + 3h + 2. Also f(1)=2(1)2−1+1=2f(1) = 2(1)^2 - 1 + 1 = 2. So the numerator is f(1+h)−f(1)=2h2+3h+2−2=2h2+3hf(1+h)-f(1) = 2h^2 + 3h + 2 - 2 = 2h^2 + 3h.

Dividing by hh gives the secant slope 2h2+3hh=2h+3\dfrac{2h^2+3h}{h} = 2h+3 (for h≠0h \neq 0). Taking the limit as h→0h \to 0 gives lim⁡h→0(2h+3)=3\displaystyle\lim_{h\to 0}(2h+3) = 3. This value, 33, is the slope of the tangent line to ff at x=1x=1, i.e., f′(1)=3f'(1) = 3.

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