limx→0log cosxx=
We have, limx→0log cosxx
At x=0,the value of the expression is the form 00
applying L'Hospital Rule
limx→0ddx(log cosx)ddxx
= limx→01cos x.ddxcosx1
= limx→01cos x∗(−sin x)
limx→0−tan x
putting x=0
=-tan 0=0.
Find:
(i)limx→2[x]
(ii)limx→52[x]
(iii)limx→1[x]
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