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EXERCISE 8.1 · Q26

Q.Which of the following functions has a removable discontinuity? If it has a removable discontinuity, redefine the function so that it becomes continuous: f(x)=(3−8x3−2x)1/xf(x) = \left(\dfrac{3-8x}{3-2x}\right)^{1/x}, for x≠0x \ne 0.

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f(x)=(3−8x3−2x)1/xf(x)=\left(\dfrac{3-8x}{3-2x}\right)^{1/x} for x≠0x\ne0; f(0)f(0) is not assigned.

As x→0x\to0, the base 3−8x3−2x→1\dfrac{3-8x}{3-2x}\to1 while the exponent 1/x→∞1/x\to\infty — a 1∞1^\infty form. Write

3−8x3−2x=1+(3−8x)−(3−2x)3−2x=1+−6x3−2x.\frac{3-8x}{3-2x}=1+\frac{(3-8x)-(3-2x)}{3-2x}=1+\frac{-6x}{3-2x}.

So with g(x)=−6x3−2xg(x)=\dfrac{-6x}{3-2x} (which →0\to0 as x→0x\to0),

lim⁡x→0(1+g(x))1/x=exp⁡(lim⁡x→0g(x)x)=exp⁡(lim⁡x→0−63−2x)=exp⁡(−63)=e−2.\lim_{x\to0}\left(1+g(x)\right)^{1/x}=\exp\left(\lim_{x\to0}\frac{g(x)}{x}\right)=\exp\left(\lim_{x\to0}\frac{-6}{3-2x}\right)=\exp\left(\frac{-6}{3}\right)=e^{-2}. …

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