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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!

  • 0
    Apart from everything else: It is absolutely forbidden to use the letter $k$ for a continuous variable.2012-04-25

1 Answers 1

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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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    @mathstude$n$t Yes, I just added that.2012-04-23