Question

Limit x tends to zero 2^x-1/e^2x-1

Answer 1

$\rm \longrightarrow \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ {2}^{x} - 1 }{ {e}^{2x} - 1 } }$

If we put x =0 in the above expression then we will get 0/0 form which is an indeterminate form. So, we will apply L'Hospital rule.

$\rm \longrightarrow \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ \dfrac{d({2}^{x} - 1 )}{dx} }{\dfrac{d( {e}^{2x} - 1 )}{dx} } }$

$\rm \longrightarrow \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ \dfrac{d({2}^{x} )}{dx} \: - \dfrac{d(1 )}{dx} }{\dfrac{d( {e}^{2x} )}{dx} - \dfrac{d(1 )}{dx}} }$

$\rm \longrightarrow \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ ({2}^{x} )ln2\: - 0 }{2 {e}^{2x} - 0} }$

$\rm \longrightarrow \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ ln2 \: \times ({2)}^{x} \: }{2 \: \times ({e)}^{2x} } }$

Now put x = 0 :-

$\rm \longrightarrow \displaystyle \large { \rm \: \dfrac{ ln2 \: \times ({2)}^{0} \: }{2 \: \times ({e)}^{0} } }$

$\rm \longrightarrow \displaystyle \large { \rm \: \dfrac{ ln2 \: \times 1 \: }{2 \: \times 1 } }$

$\rm \longrightarrow \displaystyle \large { \rm \: \dfrac{ ln2 }{2 } }$

$\rm \therefore \displaystyle \large\lim_ { \rm \: x \to \:0 }{ \rm \: \dfrac{ {2}^{x} - 1 }{ {e}^{2x} - 1 } } = \large { \rm \: \dfrac{ ln2 }{2 } }$

Additional Information :-

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x