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Exercise 10.5 · Q18

Q.It is given that f′(a)f'(a) exists, then lim⁡x→axf(a)−af(x)x−a\displaystyle\lim_{x\to a}\dfrac{xf(a)-af(x)}{x-a} is

(1) f(a)−af′(a)f(a)-af'(a)
(2) f′(a)f'(a)
(3) −f′(a)-f'(a)
(4) f(a)+af′(a)f(a)+af'(a)
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Step 1. Rewrite the numerator by adding and subtracting af(a)af(a):

xf(a)−af(x)=xf(a)−af(a)+af(a)−af(x)=f(a)(x−a)−a(f(x)−f(a))xf(a)-af(x)=xf(a)-af(a)+af(a)-af(x)=f(a)(x-a)-a\big(f(x)-f(a)\big)

Step 2. Divide by (x−a)(x-a):

xf(a)−af(x)x−a=f(a)−a⋅f(x)−f(a)x−a\frac{xf(a)-af(x)}{x-a}=f(a)-a\cdot\frac{f(x)-f(a)}{x-a}

Step 3. Take the limit as x→ax\to a. Since f′(a)f'(a) exists, lim⁡x→af(x)−f(a)x−a=f′(a)\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a) by the definition of the derivative: …

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