Just want to check this one:
I got:
lim
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!
No comments:
Post a Comment