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MISCELLANEOUS EXERCISE 9 (II) · Q74

Q.If f(2)=4f(2)=4, f′(2)=1f'(2)=1 then find lim⁡x→2[xf(2)−2f(x)x−2]\displaystyle\lim_{x\to2}\left[\dfrac{xf(2)-2f(x)}{x-2}\right]

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Given f(2)=4, f′(2)=1f(2)=4,\ f'(2)=1, find lim⁡x→2[xf(2)−2f(x)x−2]=lim⁡x→24x−2f(x)x−2\displaystyle\lim_{x\to2}\left[\dfrac{xf(2)-2f(x)}{x-2}\right]=\lim_{x\to2}\dfrac{4x-2f(x)}{x-2} (substituting f(2)=4f(2)=4).

This is a 00\frac00 form at x=2x=2 (numerator =4(2)−2f(2)=8−8=0=4(2)-2f(2)=8-8=0). Rewrite the numerator by inserting and subtracting 88:

4x−2f(x)=4x−8−2f(x)+8=4(x−2)−2[f(x)−f(2)]4x-2f(x)=4x-8-2f(x)+8=4(x-2)-2\big[f(x)-f(2)\big]

4x−2f(x)x−2=4(x−2)−2[f(x)−f(2)]x−2=4−2⋅f(x)−f(2)x−2\dfrac{4x-2f(x)}{x-2}=\dfrac{4(x-2)-2[f(x)-f(2)]}{x-2}=4-2\cdot\dfrac{f(x)-f(2)}{x-2} …

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