I'm trying to evaluate this sum of limits:

$$ \lim_{x \to 4} \frac{x^4 - 64}{x-4} + \lim_{x \to 900} \frac{900-x}{30-\sqrt{x}} $$

And I noticed that this limit $ \lim_{x \to 4} \frac{x^4 - 64}{x-4}$ doesn't exist, since the numerator is positive and the denominator is positive for $x \to 4^+$ and negative for $x \to 4^-$. But the $\lim_{x \to 900} \frac{900-x}{30-\sqrt{x}}$ exists and is equal to $60$ (I used L'Hôpital's rule). So my intuition says this sum can't exist, because I can't sum something that doesn't exist to something that exists, but the lack of rigor in this is making me suspicious, specially because Wolfram says the limit is $\infty$.

Any help would be appreciated. Thanks.

  • 1
    $\begingroup$ You are right. The limit does not exist. $\endgroup$ – Math Lover Jun 13 '18 at 17:25
  • $\begingroup$ An obvious comment, just for completeness' sake: the limit of the sums may however be well-defined even if each limit individually doesn't exist - e.g. $\lim_{x\rightarrow\infty}[(x)+(-x)]$. $\endgroup$ – Noah Schweber Jun 13 '18 at 17:37

You are almost right. The only problem is where you write that the first limit doesn't exist “since the numerator is positive and the denominator is positive for $x\to4^+$ and negative for $x\to4^−$”. What you can deduce from this is that either the limit doesn't exist or that it is equal to $0$. But you are right: the limit doesn't exist. And since the other limit exists, the sum of the limits doesn't exist either.

  • $\begingroup$ Oh, yes. When I said positive and negative I meant strictly, but you're right. Thanks. $\endgroup$ – creepyrodent Jun 13 '18 at 17:30
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    $\begingroup$ Strictly doesn't necessarily cut it. For example, $x^3$ is strictly positive for $x > 0$ and strictly negative for $x < 0$, but its limit is still $0$. $\endgroup$ – Theo Bendit Jun 13 '18 at 17:31

You can use algebra of limits. If $\lim_{x \to 4} \frac{x^4 - 64}{x - 4}$ existed, and was equal to $L$, then by the algebra of limits, $$192 = \lim_{x \to 4} x^4 - 64 = \lim_{x \to 4} \frac{x^4 - 64}{x - 4} \cdot \lim_{x \to 4} x - 4 = L \cdot 0 = 0,$$ which is a contradiction.

As for the sum, I'm not even comfortable with the expression. The sum of limits shouldn't even be written down unless the limits are known to exist (or are assumed to exist). I think the question shouldn't be, "Does this sum of limits exist?" as much as it should be "Is this expression well-defined?".

EDIT: In response to the question in the comments:

But isn't this correct? $\lim_{x \to 0} 1/x$ and $\lim_{x \to 0} -1/x$ doesn't exist, but $$ \lim_{x \to 0} \frac{1}{x} + \lim_{x \to 0} -\frac{1}{x} = \lim_{x \to 0} \frac{1}{x} - \frac{1}{x} = \lim_{x \to 0} 0 = 0?$$

No, $\lim_{x \to 0} 1/x + \lim_{x \to 0} -1/x$ does not make sense, although the other equalities are fine. In order to parse the sum, you first must take limits (both undefined), and then sum them (how!?).

Think about it: how else would you define such an expression? If you mean sum them, then take the limit, we already have an expression for this: $$\lim_{x \to 0} \frac{1}{x} - \frac{1}{x}.$$ Otherwise, we cannot sum these undefinable quantities. It becomes particularly problematic, as you've noticed, when summing two limits that approach different values. When you combined the expressions into one limit, you could at least make sense of it by tying the two variables in the two limits together. If they approach different values, what does this quantity even mean if not "find limits first, then sum them"?

  • $\begingroup$ But isn't this correct? $\lim_{x \to 0} 1/x$ and $\lim_{x \to 0} -1/x$ doesn't exist, but $\lim_{x \to 0} 1/x + \lim_{x \to 0} -1/x = \lim_{x \to 0} 1/x - 1/x = \lim_{x \to 0} 0 = 0$? $\endgroup$ – creepyrodent Jun 13 '18 at 17:45
  • $\begingroup$ @dude3221: Edited with a reply. $\endgroup$ – Theo Bendit Jun 13 '18 at 17:56

Suppose that the limit of a sum of two functions exists and the limit of one by itself also exists. Then you have a situation where for some real $L$ and $M$ $$\lim_{x \rightarrow a}f(x) + g(x) = L$$ and $$\lim_{x \rightarrow a}f(x)=M.$$ Whenever both limits exist, we can use the difference rule for limits, giving us $$L-M =\lim_{x \rightarrow a}f(x) + g(x) - \lim_{x \rightarrow a}f(x) = \lim_{x \rightarrow a}f(x) + g(x)-f(x)=\lim_{x \rightarrow a} g(x).$$ This proves the limit of the other function must exist as well.


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