Find this limit.?

Question

lim x tends to 0 of the function : {e - (1+x)^1/x}/tan x where e is the natural no. and is equal to 2.17

Sahil,

Do you study from Amit M. Agarwal...if not, how did you think of the expansion?

Answer 1

solve this by expanding the functions (1+x)^1/x and tan x

(1+x)^1/x = e[1-(x/2)+(11x^2/24).........]

tanx = x+(x^3/3)......

substitute these expansions in da limit to get

{e-e[1-(x/2).....]} / {x=(x^3/3)......}

on solving take x common from numerator and denominator and put x=0 to get the answer

answer is e/2 = 1.085

Answer 2

Undefined (0/0)

Answer 3

as x thends to zero it is (e-e)/tan x . that is 0/0

so use l'hospitals rule
denominaotr d/dx(tan x) = sec^x = 1 at x =0
numerator

let y = (1+x)^1/x

ln y = 1/x(ln(1+x)
xln y = ln (1+x)

differentiate both sides

ln y + x/ydy/dx = 1/(1+x)
or yln y + x dy/dx = 1/(1+x)
xdy/dx = 1/(1+x) - y ln y = 1/(1+x) - 1/x(in 1+ x) y
dy/dx = 1/x(1+x) - 1/....


the denomiantor is zero so the limit is infinte