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

lim_(x->0)(e^(5x) - 1)/(3x)

find lim_(x->0) (e^(x+3)-e^3)/x

Evaluate : lim_(x to 0) (e^(x)-1)/x

Evaluate lim_(xto0) (e^(x)-1-x)/(x^(2)).

If f(x) = lim_(n->oo) tan^(-1) (4n^2(1-cos(x/n))) and g(x) = lim_(n->oo) n^2/2 ln cos(2x/n) then lim_(x->0) (e^(-2g(x)) -e^(f(x)))/(x^6) equals

lim_(x->e) (lnx-1)/(x-e)

In the neighbourhood of x=0 it is known that 1+|x|lt(e^(x)-1)/(x)lt1-|x|"then find"lim_(xto0)(e^(x)-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

The value of (lim)_(x->0)(e^(x^2)-e^x+x)/(1-cos2x) is

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