What does $\lim_{x\to4} \frac{1}{x-4}$ equal? $$\lim_{x\to4} \frac{1}{x-4}$$
Would it be correct to say that the limit is undefined because the denominator would be $0$?
 A: Here is a plot of $\displaystyle f(x)=\frac{1}{x-4}$.

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.
A: 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.
A: The limit does not exist because the left limit is $-\infty$ while the right limit is $\infty$
A: 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
A: $$\lim_{x\to4^{+}} \frac{1}{x-4}= \infty$$
$$\lim_{x\to4^{-}} \frac{1}{x-4}=- \infty$$
The limit from the left is different than the one from the right so the limit doesn't exist.
A: 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:
$$\lim_{x\to 4} |f(x)| = \infty$$
