Define $f(x)$ in the following way $f(x)=\frac{(-1)^{[\frac{1}{x}]}}{x}$ where $[y]$ denotes the integer part of $y$. Then the limit $\lim_{x \to 0^+}f(x)$ does not exist: however the graph of this function splits into two "pieces" one approaching $+\infty$ and the other $-\infty$. Do we say that $f$ has got (say right, for left it is analogous) vertical asymptote?
Vertical asymptote of certain function
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calculus
real-analysis
analysis
1 Answers
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As per the definition from Wikipedia:
The line $x = a$ is a vertical asymptote of the graph of the function $y = ƒ(x)$ if at least one of the following statements is true:
$\lim _{x\to a^{-}}f(x)=\pm \infty\\ \lim _{x\to a^{+}}f(x)=\pm \infty$
Indeed, this is true, so we have a vertical asymptote.
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0But here the limit $\lim_{x \to 0^+}f(x)$ does not exist however there is no sequence $x_n \to 0^+$ such that $f(x_n)$ remains bounded. – 2017-01-20
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0@Totentanz I thought about this, and the best response I have is to take limits inferior and superior to separate the cases. – 2017-01-20