MHB Investigating Continuity of $f(x)

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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?
 
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Alexmahone said:
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. \cos\left(\frac1x\right) is oscillating between -1 and 1 if x approaches zero.

2. \lim_{x \to 0}(\sqrt{x}) = 0

3. Therefore
\lim_{x \to 0} \left(\sqrt{x}\cos\left(\frac{1}{x}\right) \right) = 0
 
earboth said:
1. \cos\left(\frac1x\right) is oscillating between -1 and 1 if x approaches zero.

2. \lim_{x \to 0}(\sqrt{x}) = 0

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

Thanks, but you didn't answer my question. I asked if the function is continuous at 0.
 
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.
 
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