A Limiting Piecewise Function

Calculus Level 3

$\large \lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x }{ a } \left\lfloor \frac { b }{ x } \right\rfloor }$ Compute the limit above for constants $a, b >0$ .

\frac{b}{a}

\frac{a}{b}

0 Does not exist

1 solution

Carlos Herrera
Mar 16, 2016

It is not hard to see that: $b-x<x\left\lfloor \frac { b }{ x } \right\rfloor<b$ , for all $x>0$

But, $(b-x ) \rightarrow b$ as $x \rightarrow {0}^{+}$ and $b \rightarrow b$ as $x \rightarrow {0}^{+}$ , so $\large \lim _{ x\rightarrow { 0 }^{ + } }{ x\left\lfloor \frac { b }{ x } \right\rfloor }=b$ We can conclude that: $\large \lim _{ x\rightarrow { 0 }^{ + } }{ \frac { x }{ a } \left\lfloor \frac { b }{ x } \right\rfloor }=\frac{b}{a}$

Great solution, Carlos! Since your solution uses the squeeze theorem, I might recommend including something like

"By the squeeze theorem , we have that..."

In order to get that link to appear, you can use

[[wiki-squeeze-theorem|squeeze theorem]]

Andrew Ellinor - 5 years, 2 months ago

Thanks :).

Carlos Herrera - 5 years, 2 months ago

A Limiting Piecewise Function

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