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
please help me understand why the following limit does not exist?
Best Answer
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).