Find a limit of function: $\lim_{x\to 0}\frac{(-2)^{\lfloor x \rfloor}}{x} $

Question

Find $\displaystyle \lim_{x\to 0}\left(\frac{(-2)^{\lfloor x \rfloor}}{x} \right)$

$\lfloor x \rfloor$ - is the floor function

From graphic it seems that this limit is different from right and left.

Fly by Night · Accepted Answer · 2016-11-12T23:02:52.797

1

HINT:

$\lfloor x \rfloor = -1$ for all $-1 < x < 0$ and $\lfloor x \rfloor = 0$ for all $0 < x < 1$.

To find the limit $x \to 0^-$, replace $\lfloor x \rfloor$ with $-1$, and then let $x \to 0^-$ ($x < 0$ and $x \to 0$)

To find the limit $x \to 0^+$, replace $\lfloor x \rfloor$ with $0$, and then let $x \to 0^+$ ($x > 0$ and $x \to 0$)

For the limit $x \to 0$ to exist, we need the limit $x \to 0^-$ and the limit $x \to 0^+$ to be equal and finite.

edited Nov 12 '16 at 23:02

answered Nov 12 '16 at 22:36

Fly by Night

32,272

$\displaystyle \lim_{x\to 0^{+}}\left(\frac{(-2)^{\lfloor x \rfloor}}{x} \right)$ = \infty – Nowak Grzegorz Nov 12 '16 at 22:41
@NowakGrzegorz So the limit does not exist – Fly by Night Nov 12 '16 at 22:43
@NowakGrzegorz \infty will give $\infty$ – Fly by Night Nov 12 '16 at 22:45
$\displaystyle \lim_{x\to 0^{-}}\left(\frac{(-2)^{\lfloor x \rfloor}}{x} \right)$ = 0 – Nowak Grzegorz Nov 12 '16 at 22:45
@NowakGrzegorz If $-1 < x<0$ then $\lfloor x \rfloor = -1$ so
$$\frac{(-2)^{\lfloor x \rfloor}}{x} = - \frac{1}{2x}$$

Since $x<0$, $ - \frac{1}{2x} > 0$ and as $x \to 0^-$, $ - \frac{1}{2x} \to + \infty$
– Fly by Night Nov 12 '16 at 22:52
@NowakGrzegorz You need the \infty to be between the dollar signs. – Fly by Night Nov 12 '16 at 22:53
I know $\infty$ but I can't edit after 5 minuts – Nowak Grzegorz Nov 12 '16 at 22:56
THE ANSWER: $\displaystyle \lim_{x\to 0}\left(\frac{(-2)^{\lfloor x \rfloor}}{x} \right)$ = $+ \infty$ Is it ok? – Nowak Grzegorz Nov 12 '16 at 22:59
@NowakGrzegorz So we say that the limit does not exist. The limit needs to be finite for it to be said to exist. We can say that $\mathrm f(x) \to \infty$ as $x \to 0$. But we say that $\lim_{x \to 0} \mathrm f(x)$ is undefined. – Fly by Night Nov 12 '16 at 22:59

score 1 · Answer 2 · answered Nov 12 '16 at 23:28

Since it is not restrictive to assume your function is defined over $(-1,1)\setminus\{0\}$, we have $$ \lfloor x\rfloor=\begin{cases} -1 & \text{if $-1<x<0$} \\[4px] 0 & \text{if $0<x<1$} \end{cases} $$ Thus $$ \lim_{x\to0^+}\frac{(-2)^{\lfloor x\rfloor}}{x}= \lim_{x\to0^+}\frac{1}{x}=+\infty $$ Also $$ \lim_{x\to0^-}\frac{(-2)^{\lfloor x\rfloor}}{x}= \lim_{x\to0^-}\frac{-1/2}{x}=+\infty $$ Thus $$ \lim_{x\to0}\frac{(-2)^{\lfloor x\rfloor}}{x}=+\infty $$

Find a limit of function: $\lim_{x\to 0}\frac{(-2)^{\lfloor x \rfloor}}{x} $

2 Answers2