Question

c). If f(x)={x*(sin|x|/x) for x is not equal to zero
{0 for x=0
, then discuss continuity of f(x) at x=0.

Answer 1

if the limit as x appraoches 0 and (f0) is the same then it is continuous:
\[\lim_{x \rightarrow 0}=f(0)?\]

Answer 2

plz write complete solution step by step

Answer 3

\[\lim_{x \rightarrow 0} \frac{xsin|x|}{x}=f(0)?\]
f(0)=0 from ur problem so:
\[\lim_{x \rightarrow 0}\frac{xsin|x|}{x}=0?\]
the x cancels out so:
\[\lim_{x \rightarrow 0}\sin|x|=\sin(0)=0\]
and
0=0
so then the limit does equal to point so it is continuous

Answer 4

plz look at the question attached

Answer 5

ok u have to use squeeze theorem

Answer 6

u kno that sin of any angle has to b between -1 and 1 right?
so:
-1<sin(any angle)<1

Answer 7

\[-1 \le \sin(anyangle) \le 1\]

Answer 8

in this case the angle is:
\[\frac{|x|}{x}\]
so
\[-1 \le \sin(\frac{|x|}{x}) \le 1\]
now lets multiply each side by x so it becomes:
\[-x \le x*\sin(\frac{|x|}{x}) \le x\]
now the middle is the limit u want to solve. so lets solve the left and right side for the limit we want (0)
\[\lim_{x \rightarrow 0} -x \le x*\sin(\frac{|x|}{x}) \le x\]
becomes
\[0 \le x*\sin(\frac{|x|}{x}) \le 0\]
so it must be greater than or equal to 0 and less than or equal to 0 and the only value that can fulfil both of these is 0 so it must equal 0
so ur limit and point are equal and it is continuous

Answer 9

\[\lim_{x \rightarrow 0}x*\sin(\frac{|x|}{x})=0\]

Answer 10

Yo