What is $\lim\limits_{k \to 0}{f(k) = 2 + k^{\frac{3}{2}}\cos {\frac{1}{k^2}}}$ 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!

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