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- Feb 15, 2012

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checking for continuity is a good idea, but in this particular problem it doesn't rule out any points for us, f is continuous everywhere:

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2 = 2$$

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{2}{x^2} = 2$$

so f is continuous at 1.

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2}{x^2} = \frac{2}{4} = \frac{1}{2}$$

$$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$$

so f is continuous at 2.

there's no problem at 0, f(0) = 2.

instead, you should look at places where the derivative might not exist.

$$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 2 = 2$$

$$\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{2}{x^2} = 2$$

so f is continuous at 1.

$$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{2}{x^2} = \frac{2}{4} = \frac{1}{2}$$

$$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \frac{1}{x} = \frac{1}{2}$$

so f is continuous at 2.

there's no problem at 0, f(0) = 2.

instead, you should look at places where the derivative might not exist.

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- Jan 29, 2012

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Since the question asks about "differentiability" I would think that the obvious first thing to do would be to look up the definition of "differentiable". Fortunately, for a function of a single variable that is simply "has a derivative there" (for functions of more than one variable it is more complicated).Hello,

How do I solve this kind of problems ?

For which values of x the next function is "differentiable" ?

View attachment 530

I know it has something to do with the existent of the one sided limits, but which limits should I be calculating exactly ?

Thanks !

So what is the

(You are completely wrong when you say the function is not continuous. I think you need to review the basic definitions.)

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