limx→0√1+x2−√1−x2x
Rationalising the numenator
limx→0(√1+x2−√1−x2)x×(√1+x2+√1−x2)(√1+x2+√1−x2)
=limx→0(1+x2)−(1−x2)(√1+x2+√1−x2)
=limx→02x2(√1+x2+√1−x2)
=2×0(√1+√1)
=22×0=0
Evaluate the following one sided limits:
(i)limx→2+x−3x2−4
(ii)limx→2−x−3x2−4
(iii)limx→0+13x
(iv)limx→8+2xx+8
(v)limx→0+2x15
(vi)limx→π−2tan x
(vii)limx→π2+sec x
(viii)limx→0−x2−3x+2x3−2x2
(ix)limx→−2+x2−12x+4
(x)limx→0+(2−cot x)
(xi)limx→0−1+cosecx
limx→0√1+x+x2−√x+12x2