A limit

Calculus Level 4

Consider the following limits

$\large L_- = \lim_{x \to 0^-} (1+\{x\})^{\lfloor x \rfloor} \qquad \qquad L_+ = \lim_{x \to 0^+} (1+\{x\})^{\lfloor x \rfloor}$

What are $L_-$ and $L_+$ ?

Notations:

$\lfloor \cdot \rfloor$ denotes floor function
$\{ \cdot \}$ denotes fractional part function

L_-=0.5; \ L_+=1

L_-.=-1 ; \ L_+=-0.5

L_-=-1 ; \ L_+=-1

L_-=-0.5 ; \ L_+=1

L_-=0.5 ; \ L_+=-1

1 solution

Aditya Narayan Sharma
Apr 25, 2017

$\displaystyle \begin{aligned} L_{-}&=\lim_{x\to 0^{-}} \left(1+\{x\}\right)^{\lfloor{x\rfloor}} \\ &= \lim_{x\to 0} \left(1+x-(-1)\right)^{-1}\quad\text{Since }\lfloor{x\rfloor}=-1\quad\text{for }x\in(-1,0) \\ &= \left(1+1 \right)^{-1} \\ &=0.5 \end{aligned}$

Similarly,

$\displaystyle \begin{aligned} L_{+}&=\lim_{x\to 0^{+}} \left(1+\{x\}\right)^{\lfloor{x\rfloor}} \\ &= \lim_{x\to 0} \left(1+x-(0)\right)^{0}\quad\text{Since }\lfloor{x\rfloor}=0\quad\text{for }x\in(0,1) \\ &= \left(1+0\right)^{0} \\ &=1 \end{aligned}$

Hence LHL= $0.5$ and RHL= $1$

A limit

1 solution

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