1) is 0
2) is |sinx|x
lim x-->0+ = 1
lim x-->0- = -1
l.h.l≠r.h.l
so limit does not exist
1)) \lim_{x\rightarrow \propto } sinx/x
2) \lim_{x\rightarrow 0} \frac{\sqrt{(1-cos2x)/2}}{x}
1) is 0
2) is |sinx|x
lim x-->0+ = 1
lim x-->0- = -1
l.h.l≠r.h.l
so limit does not exist
1)
using sandwich theorem
-1/x ≤ (sin x)/x ≤ 1/x
Now as → infinity we know that both 1/x and -1/x tend to 0,
so (sin x)/x must also tend to 0, by the sandwich theorem