Investigating Continuity of $f(x)

[alexmahone]

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?

 

[earboth]

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

 

[alexmahone]

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.

 

[Ackbach]

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.