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Suppose the function $f(x)$ has the properties: $\lim_{x\to 1^+}f(x) = 1$ and $\lim_{x\to 1^-}f(x) = -2$. It follows that $\lim_{x\to 1}f(x)$ does not exist.

Another function $h(x)$ has the properties: $\lim_{x\to 1^+}h(x) = -2$ and $\lim_{x\to 1^-}h(x) = 1$. It follows that $\lim_{x\to 1}h(x)$ does not exist.

1. $$\lim_{x\to 1^-}[f(x)+h(x)] = \lim_{x\to 1^-}f(x) + \lim_{x\to 1^-}h(x) = -2 + 1 = -1$$

2. $$\lim_{x\to 1^+}[f(x)+h(x)] = \lim_{x\to 1^+}f(x) + \lim_{x\to 1^+}h(x) = 1 + (-2) = -1$$

3. $$\lim_{x\to 1^-}[f(x)+h(x)] = \lim_{x\to 1^+}[f(x)+h(x)] = -1$$ $$\lim_{x\to 1}[f(x)+h(x)]=-1$$

Also, $$\lim_{x\to 1}[f(x)+h(x)] = \lim_{x\to 1}f(x) + \lim_{x\to 1}h(x)$$

However, the left hand side of this equation is equal to $-1$ while the right hand side of this equation is not defined; the right hand side of the equation is the sum of two undefined values.

My question regarding this problem is:

Is it true to say that $$\lim_{x\to 1}[f(x)+h(x)]=-1?$$ Or, is the problem a case where the addition rule for limits do not hold?

Bérénice
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Cyclopropane
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    Yes, $\lim_{x\to 1}(f(x)+h(x))=-1$ The addition rule holds, but it requires that the individual limits exist. The individual limits do not exist here and it is incorrect, obviously, to write $\lim_{x\to 1}(f(x)+h(x))=\lim_{x\to 1}f(x)+\lim_{x\to 1}h(x)$ since the right-hand side fails to exist. – Mark Viola Apr 25 '17 at 18:59
  • (+1) because it's rare to see well redacted questions for new users. – zwim Apr 25 '17 at 19:15

1 Answers1

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The theorem is:

If $\lim_{x \to a} f(x)$ and $\lim_{x \to a}g(x)$ exist, then $\lim_{x > \to a} (f(x)+g(x))$ exists, and equals $\lim_{x \to a}f(x) +\lim_{x > \to a} g(x)$

There is no such theorem in the case that $\lim f$ or $\lim g$ do not exist, because as you note the RHS does not make any sense in those circumstances.

preferred_anon
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