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sineoko [7]
3 years ago
12

Find the values of x (if any) at which f(x)= x+1/x^2-1 is not continuous. If so, is the discontinuity removable?

Mathematics
1 answer:
Mkey [24]3 years ago
8 0

f(x)=\dfrac{x+1}{x^2-1} has two points of discontinuity at x=\pm1, where x^2-1=0.

We have

\dfrac{x+1}{x^2-1}=\dfrac{x+1}{(x+1)(x-1)}

so as long as x\neq-1, we can cancel x+1 and be left with \dfrac1{x-1}. Then as x\to-1, we have

\displaystyle\lim_{x\to-1}f(x)=\lim_{x\to-1}\frac1{x-1}=-\frac12

The limit exists, so the discontinuity at -1 is removable (so definitely not C or D).

Meanwhile,

\displaystyle\lim_{x\to1}f(x)=\lim_{x\to1}\frac1{x-1}

does not exist, so the discontinuity at 1 is non-removable, making A the correct answer.

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