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$$\lim_{x\to4} \frac{1}{x-4}$$ Would it be correct to say that the limit is undefined because the denominator would be $0$?

Ian Mateus
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    No, but it would be correct to say that the limit is undefined because the two-sided limit doesn't exist. – rae306 Sep 06 '14 at 18:18
  • What about saying that the limit does not exist because there is a vertical asymptote at x=4? –  Sep 06 '14 at 18:19
  • $\lim_{x\to4^+}\frac{1}{x-4}=+\infty\neq-\infty=\lim_{x\to4^-}\frac{1}{x-4}$ – barak manos Sep 06 '14 at 18:20
  • consider below limit : $$\lim \limits_{x \to 4} \dfrac{1}{(x-4)^2}$$ the left and right limits are $+ \infty$, if your textbook defines this as "exists", then the limit exists.(Notice that $x=4$ is still an asymptote) – AgentS Sep 06 '14 at 18:28

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Here is a plot of $\displaystyle f(x)=\frac{1}{x-4}$.

enter image description here

As you can see, $\displaystyle \lim_{x\to4^-}=-\infty$ (the limit as $x$ approches $4$ from the left), whereas $\displaystyle \lim_{x\to4^+}=\infty$ (the limit as $x$ approaches $4$ from the right).

Therefore, we say that the limit doesn't exist.

rae306
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The limit does not exist, but not for the reason you give.

Recall that solving limits often involves more than simply "plugging in" the value that $x$ approaches. What is happening with limit such as this is that $x$ is getting arbitrarily close to $4$, not what happens exactly at $4$.

We need with this function, to see what is happening to the value of the function as $x$ approaches $4$ from the left, and also what happens as $x\to 4$ from the right.

The closer it gets, from the left, $\frac 1{x-4}$ blows up, in the negative direction. The closer it gets, from the right, $\frac 1{x-4}$ blows up, in the positive direction.

Indeed, as $$\lim_{x\to 4^+} \frac 1{x-4} \to +\infty$$

$$\lim_{x\to 4^-} \frac 1{x-4} \to -\infty$$

Since the one-sided limits disagree, the limit does not exist.

amWhy
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  • Would it be correct to say that the limit does not exist because there is an asymptote at x=4? –  Sep 06 '14 at 18:25
  • Recall that evaluating a limit of a function as $x \to a$ isn't equivalent to evaluating the function at $x = a$. Sometimes it is that simple, but not always. But indeed, the graph of this function and its asymptote gives us the information of what happens to the function value as $x \to 4$ from each side. – amWhy Sep 06 '14 at 18:31
  • If, on the other hand, we were examining $\lim_{x\to 4}\frac 1{(x - 4)^2}$, we find that there is an asymptote, but the value of the function as $x$ approaches $4$ from both the left and the right blows up, in the positive direction, so we say this limit approaches $+ \infty$ as $x \to 4$. – amWhy Sep 06 '14 at 18:38
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The limit does not exist because the left limit is $-\infty$ while the right limit is $\infty$

lovelesswang
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Hint: The limit does not exist, the one sided limits does not exist either and while on one side the function diverges to infinity, on the other side it diverges to minus infinity

See this graph for illustration

Belgi
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$$\lim_{x\to4^{+}} \frac{1}{x-4}= \infty$$ $$\lim_{x\to4^{-}} \frac{1}{x-4}=- \infty$$ The limit from the left is different than the one from the right so the limit doesn't exist.

UserX
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Let $f$ be a real-valued function on $\mathbb R$. (To be absolutely correct: Let $f$ be a real-valued function on $(a-h,a) \cup (a,a+h)$ for some $h > 0$.) If $a \in \mathbb R$, then the statement $$\lim_{x\to a} f(x) = b$$ means: For all $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(x)-b| < \varepsilon$ whenever $|x-a| < \delta$.

I expect you know this.

However, the statement

$$\lim_{x\to a} f(x) = \infty$$

means something entirely different! It doesn't mean:

For all $\varepsilon > 0$ there exists $\delta > 0$ such that $|f(x)-\infty| < \varepsilon$ whenever $|x-a| < \delta$.

That would make no sense! Instead, it means this:

For all $M > 0$ there exists $\delta > 0$ such that $f(x) > M$ whenever $|x-a| < \delta$.

You can see the formal similarities between the two statements (I could even have used $\varepsilon$ instead of $M$ in the second statement, but $\varepsilon$ has been typecast as a very small number, and is not accustomed to playing bigger roles). But they mean completely different things.

So now to your question: The function $f(x) = \dfrac{1}{x-4}$ satisfies none of the above, because $f(x)$ attains arbitrarily large positive and negative values arbitrarily close to $x=4$. So $\lim_{x\to 4} f(x)$ does not exist. What we can say is this:

$$\lim_{x\to 4} |f(x)| = \infty$$

TonyK
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