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Tangent line to f(x) = 1/x at x = 2 using the limit definition

Consider the function f(x)=1xf(x) = \dfrac{1}{x}.

Use the limit definition of the derivative, f′(a)=lim⁡h→0f(a+h)−f(a)h,f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}, to find f′(2)f'(2) for f(x)=1xf(x) = \dfrac{1}{x}. Then use this result to find the equation of the tangent line to the graph of ff at x=2x = 2.

Hint

Write out f(2+h)f(2+h) and f(2)f(2) as fractions, combine them into a single fraction over a common denominator, simplify the numerator, then cancel the hh before letting h→0h \to 0. Once you have f′(2)f'(2), use the point (2,f(2))(2, f(2)) and point-slope form to write the tangent line.

Step-by-step solution

Compute f(2+h)−f(2)=12+h−12=2−(2+h)2(2+h)=−h2(2+h)f(2+h) - f(2) = \dfrac{1}{2+h} - \dfrac{1}{2} = \dfrac{2 - (2+h)}{2(2+h)} = \dfrac{-h}{2(2+h)}. Dividing by hh gives f(2+h)−f(2)h=−12(2+h)\dfrac{f(2+h)-f(2)}{h} = \dfrac{-1}{2(2+h)}, since the hh in the numerator cancels with the hh in the denominator of the difference quotient.

Taking the limit as h→0h \to 0: f′(2)=lim⁡h→0−12(2+h)=−14f'(2) = \lim_{h\to 0} \dfrac{-1}{2(2+h)} = \dfrac{-1}{4}. Since f(2)=12f(2) = \dfrac{1}{2}, the tangent line passes through (2,12)(2, \tfrac{1}{2}) with slope −14-\tfrac14. Using point-slope form, y−12=−14(x−2)y - \tfrac12 = -\tfrac14(x-2), which simplifies to y=−14x+1y = -\tfrac14 x + 1.

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