Question

lim x → 0 (e^x-1)/2x​

Answer 1

lim x → 0 (e^x - 1 ) / 2x

Dividing numerator and denominator by x

lim x → 0 [(e^x - 1)/x] / 2x/x

As per logrithmic limit : lim x → 0 (e^x - 1)/x = 1

= 1/2 is Answer.

Answer 2

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \frac{ {e}^{x} - 1 }{2x}$

Now put x = 0 in the above expression.

$\bf \dfrac{ {e}^{0} - 1 }{2 \times 0}$

$\bf \dfrac{ {0}}{ 0}$

Therefore we get 0/0 form. So we can apply L'Hospital's rule to solve given problem.

In L'Hospital's rule we differentiate numerator and denominator seperately.

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{ \dfrac{d({e}^{x} - 1)}{dx} }{ \dfrac{d(2x)}{dx} }$

We know that :-

$\boxed{ \bf\dfrac{d({e}^{t} )}{dt} = {e}^{t}}$

$\boxed{ \bf\dfrac{d(c)}{dt} = 0} \bf \: where \: c \: is \: constant$

Therefore :-

$\bf \: \dfrac{d({e}^{x} - 1)}{dx} = {e}^{x}$

We also know that :-

$\boxed{ \bf\dfrac{d({x}^{t} )}{dt} = t \: {x}^{t - 1}}$

Therefore :-

$\bf \: \dfrac{d(2x)}{dx} = 2$

so we finally get :-

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{ {{e}^{x} } }{2}$

Now put x = 0

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{ {{e}^{0} } }{2}$

$\implies\displaystyle \sf \dfrac{1}{2}$

Answer :

$\displaystyle \sf \lim \limits_{x \to \: 0} \: \frac{ {e}^{x} - 1 }{2x} = \dfrac{1}{2}$