Question

Prove the following limit

lim xsin1/x = 0
x-->0

Answer 1

\[\lim_{x\rightarrow0}[x\sin(\frac{1}{x})]\]?

Answer 2

-1<=sin(1/x)<=1
-x<=xsin(1/x)<=x ( or vice versa)
run the limit

Answer 3

\(\bf \lim_{x\rightarrow0} x sin\left(\frac{1}{x}\right) = 0\\
-1 < sin\left(\frac{1}{x}\right) < 1 \implies
-x < xsin\left(\frac{1}{x}\right) < x\\
\color{blue}{\lim_{x\rightarrow0} x =0 \qquad \lim_{x\rightarrow0} -x =0}\\
\textit{by the squeeze theorem then }\\
\lim_{x\rightarrow0} x sin\left(\frac{1}{x}\right) = 0\)

Answer 4

ahemm, well, or equals, one sec

Answer 5

you need to show both sides
so
-x<=xsin(1/x)<=x
and
-x>=xsin(1/x)>=x
we have the same limit as x approaches 0 from the left and right

Answer 6

\(\bf \lim_{x\rightarrow0} x sin\left(\frac{1}{x}\right) = 0\\
-1 \le sin\left(\frac{1}{x}\right) \le 1 \implies
-x \le xsin\left(\frac{1}{x}\right) \le x\\
\color{blue}{\lim_{x\rightarrow0} x =0 \qquad \lim_{x\rightarrow0} -x =0}\\
\textit{by the squeeze theorem then }\\
\lim_{x\rightarrow0} x sin\left(\frac{1}{x}\right) = 0\)

Answer 7

now consider x negative @jdoe0001 (its trivial, but still.)

Answer 8

right

Answer 9

or by definition
\[given \space\epsilon>0\\let\ \space \delta=\epsilon\\then\\|x-0|< \delta\implies\space|x*\sin(\frac{1}{x})|\le |x|< \delta = \epsilon\]
thus it converges to 0 by definition of limits.

Answer 10

@TheDoctorRules does this make sense?