Question

limit x tends to 0 e^x-1/tanx​

Answer 1

Answer:

So, lim(x-->0) (tan x/x)^(1/x) is of the form "1^∞".

We evaluate this using logarithms.

Let L = lim(x-->0) (tan x/x)^(1/x).

So, ln L = lim(x-->0) ln(tan x/x) / x, now of the form "0/0".

Applying L'Hopital's Rule, we obtain

ln L = lim(x-->0) [sec^2(x)/tan x - 1/x] / 1

since (d/dx) ln(tan x/x) = (d/dx) [ln(tan x) - ln x] = sec^2(x)/tan x - 1/x.

Rewriting this (as this is of the form "∞ - ∞"):

ln L = lim(x-->0) sec^2(x)/tan x - 1/x

     = lim(x-->0) 1/(sin x cos x) - 1/x

     = lim(x-->0) 2/sin(2x) - 1/x, still of the form "∞ - ∞"

     = lim(x-->0) [2x - sin(2x)] / (x sin(2x)), now of the form "0/0"

Apply L'Hopital's Rule again (two times):

ln L = lim(x-->0) [2 - 2 cos(2x)] / (sin(2x) + 2x cos(2x)), still of the form "0/0"

     = lim(x-->0) 4 sin(2x) / (4 cos(2x) - 4x sin(2x))

     = 0/(4 - 0)

     = 0.

Hence, L = e^0 = 1.