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Continuity

Alexmahone

Active member
Jan 26, 2012
268
Is the function

$f(x) = \left\{\begin{array}{rcl}\sqrt{x}\cos\left(\frac{1}{x}\right)&\text{if}&x\neq 0\\0 &\text{if}&x=0\end{array}\right.$

continuous at 0?

My answer is no, because the left hand limit does not exist. Am I right?
 
Last edited:

earboth

Active member
Jan 30, 2012
74
Is the function

$f(x) = \left\{\begin{array}{rcl}\sqrt{x}\cos\left(\frac{1}{x}\right)&\text{if}&x\neq 0\\0 &\text{if}&x=0\end{array}\right.$

continuous at 0?

My answer is no, because the left hand limit does not exist. Am I right?
1. [tex]\cos\left(\frac1x\right)[/tex] is oscillating between -1 and 1 if x approaches zero.

2. [tex]\lim_{x \to 0}(\sqrt{x}) = 0 [/tex]

3. Therefore
[tex] \lim_{x \to 0} \left(\sqrt{x}\cos\left(\frac{1}{x}\right) \right) = 0[/tex]
 

Alexmahone

Active member
Jan 26, 2012
268
1. [tex]\cos\left(\frac1x\right)[/tex] is oscillating between -1 and 1 if x approaches zero.

2. [tex]\lim_{x \to 0}(\sqrt{x}) = 0 [/tex]

3. Therefore
[tex] \lim_{x \to 0} \left(\sqrt{x}\cos\left(\frac{1}{x}\right) \right) = 0[/tex]
Thanks, but you didn't answer my question. I asked if the function is continuous at 0.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
It is continuous from the right at zero. But, since the negative numbers are not in the domain (I'm assuming you're talking about a real-valued function only), it cannot be fully continuous at zero.