# Evaluating limits if it exist for one sided limits

Firstly, i think i am confused with the concept of one sided limits, in my mind i always understood one sided limits as the limits that ALWAYS exist because they are from one side, there is nothing relative to it(something relative to a one sided limit from the$^+$ is the one sided limit from the$^-$)

so i encountered the following question

Evaluate the limit, if it exists: $$\lim\limits_{x \rightarrow 1^+}{{-5\over 1-x}}$$

and apparently the answer is that it Does Not Exist

• $\pm\infty$ are not numbers. – Hirshy Sep 2 '17 at 7:15
• Infinite limits aren't limits in the strict sense of the definition of limits (that for every $\varepsilon>0$ there exists $\delta>0$ etc.). When I write something like $\lim\limits_{x\to a}f(x)=\infty$ I'd prefer to use the term "strictly divergent" to not confuse it with convergence in some weird way. It means that $f(x)$ will get arbitrarily large in a neighborhood of $a$, but it will not "converge to infinity". – Hirshy Sep 2 '17 at 7:34
As mentioned in the comments, $\pm \infty$ aren't real numbers, so by definition the limit doesn't exist. Another example is: $$\lim_{x \to 0^+} \sin(\tfrac{1}{x})$$ As $x$ approaches $0$ from the right, $\frac{1}{x}$ approaches $\infty$, and so the sine function will oscillate between $-1$ and $1$ infinitely many times, no matter how close $x$ is to $0$ from the right. Even if $\pm \infty$ were real numbers, the limit would still not exist. The graph of $f(x) = \sin(\tfrac{1}{x})$ is discontinuous at $x = 0$, even though the discontinuity is not an asymptote or a hole (point of discontinuity).
• That was what you were just told. When you read "$\lim_{x\to a} f(x)= \infty$" that is just saying that "the limit does not exist" (in a particular way). – user247327 Apr 26 '18 at 23:11