`lim_(x->0) {a^(2x)-1}/{x}`=

Use formula lim_(x->0)(a^x-1)/x=log(a) to find lim_(x->0)(2^x-1)/((1+x)^(1/2)-1)

lim_(x→0) (√(x+1)−1)/x

Let lim_(x to 0) ("sin" 2X)/(tan ((x)/(2))) = L, and lim_(x to 0) (e^(2x) - 1)/(x) = L_(2) then the value of L_(1)L_(2) is

Statement-1 : lim_(x to 0) (1 - cos x)/(x(2^(x) - 1)) = (1)/(2) log_(2) e . Statement : lim_(x to 0) ("sin" x)/(x) = 1 , lim_(x to 0) (a^(x) - 1)/(x) = log a, a gt 0

lim_(x->0) ((1+x)^(1/x)-e)/x is equal to

Evaluate: lim_(x->0) (1-cos2x)/(x^2)

Evaluate lim_(x to 0) (2^(x)-1)/((1+x)^(1//2)-1)

Evaluate the following limit: (lim)_(x->0)(cos2x-1)/(cos x-1)

Evaluate: lim_(x->0)(1/(x^2)-1/(sin^2x))

If f(x) be a cubic polynomial and lim_(x->0)(sin^2x)/(f(x))=1/3 then f(1) can not be equal to :