Sunday, November 25, 2018

Find limit $xrightarrow0$ of $f(x)=x^2cdotleft({sin{frac 1 x}}right)^2$



I have following function:



$$f(x)=x^2\cdot\left({\sin{\frac 1 x}}\right)^2$$



I want to find the limit of the function for $x\rightarrow0^\pm$. First I analyze $\frac 1 x$:




  • $\frac {1}{x}\rightarrow +\infty$ for $x\rightarrow0^+$




but the $\sin$ of infinity does not exist. Then I use the comparison theorem (I don't know how it's called in English) and conclude that, because



$$\left|{x^2\left({\sin{\frac 1 x}}\right)}^2 \right| \le \frac{1}{x^2}\rightarrow0^+$$



therefore the initial function tends to $0$. Is this reasoning correct? Are there better ways?


Answer



If you meant $\left\lvert x^2\sin^2\left(\frac1x\right)\right\rvert\leqslant x^2$, then yes, it is correct. It follows from this that $\lim_{x\to 0}\left\lvert x^2\sin^2\left(\frac1x\right)\right\rvert=0$ and that therefore $\lim_{x\to 0}x^2\sin^2\left(\frac1x\right)=0$.


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...