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Just want to check this one:

I got:

$$\displaystyle \lim_{k \to 0}{f(k) = 2} \;+\; \lim_{k \to 0}{k^{\frac{3}{2}}\cos {\frac{1}{k^2}}}$$

Since $\lim\limits_{k \to 0}\cos{\frac{1}{k^2}} = 0$, using the squeeze theorem, I have $\lim\limits_{k \to 0} k^{\frac{3}{2}}\cos{\frac{1}{k^2}} = 0$.

So
$$\begin{align*} \lim_{k \to 0}f(k) &= 2 + \lim_{k \to 0}k^{\frac{3}{2}}\cos\left(\frac{1}{k^2}\right)\\ &= 2 + 0\\ &= 2 \end{align*}$$

Is this correct?

Thanks!

  • 2
    Sorry, but it is false that $\lim\limits_{k\to 0}\cos\frac{1}{k^2}=0$. That limit does not exist: we can find $k$ arbitrarily close to $0$ where the cosine is equal to $0$, to $1$, or to $-1$. It *is* true that $\lim\limits_{k\to 0}k^{3/2}\cos(1/k^2)=0$, and this can be shown using the Squeeze Theorem, but the limit of the cosine alone does not exist.2012-04-23
  • 1
    @mathstudent Arturo is right (listen to the master). The limit of $cos\left(\frac{1}{k^2}\right)$ does not exist. http://snipurl.com/236mdru2012-04-23
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    Apart from everything else: It is absolutely forbidden to use the letter $k$ for a continuous variable.2012-04-25

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