Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\lim\limits_{k \to 0} k^{\frac{3}{2}}\cos{\frac{1}{k^2}} = 0$.

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

Is this correct?


share|cite|improve this question
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
@mathstudent Arturo is right (listen to the master). The limit of $cos\left(\frac{1}{k^2}\right)$ does not exist. – 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

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

share|cite|improve this answer
but the 2 part at the front? Wouldn't that be 2 + 0 = 2, altogether? – JackReacher Apr 23 '12 at 3:53
@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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.