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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!

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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. –  Arturo Magidin Apr 23 '12 at 4:13
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@mathstudent Arturo is right (listen to the master). The limit of $cos\left(\frac{1}{k^2}\right)$ does not exist. snipurl.com/236mdru –  Kirthi Raman Apr 23 '12 at 11:23
    
Apart from everything else: It is absolutely forbidden to use the letter $k$ for a continuous variable. –  Christian Blatter Apr 25 '12 at 16:28

1 Answer 1

Almost. Since $\lim\limits_{k\rightarrow 0^+} k^{3/2}=0$ (note the one-sided limit) and since $-1\le \cos(x)\le1$ for all $x$, it follows from the Squeeze Theorem that $\lim\limits_{k\rightarrow 0^+} \bigl[\,k^{3/2}\cos(1/k^2)\,\bigr]=0$.

Thus, $\lim\limits_{k\rightarrow 0^+} \bigl[2+k^{3/2}\cos(1/k^2)\,\bigr]=2+0=2$.

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but the 2 part at the front? Wouldn't that be 2 + 0 = 2, altogether? –  JackReacher Apr 23 '12 at 3:53
1  
@mathstudent Yes, I just added that. –  David Mitra Apr 23 '12 at 3:54
    
Thanks so much David. –  JackReacher Apr 23 '12 at 3:56

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