Question

Lim x_>0 e^sinx-1/log(1+x)

Answer 1

$\Huge{\underline{\underline{\mathfrak{Question \colon }}}}$

Evaluate the given limit :

$\lim \: \: \: \: \: \: \: \tt{\frac{ {e}^{sin \: x} - 1}{ log(1 + x) } } \: \\ \tt{x \longrightarrow}0$

$\Huge{\underline{\underline{\mathfrak{Solution \colon }}}}$

Given Expression,

$\lim \: \: \: \: \: \: \: \tt{\frac{ {e}^{sin \: x} - 1}{ log(1 + x) } } \: \\ \tt{x \longrightarrow}0$

• At x = 0,the given function is in indeterminable form.

On substituting x = 0,we would get log(1) in the denominator of the given function. But log(1) = 0,thus the given function is in n/0 form

L'Hospital Rule

• The derivatives of numerator and denominator are found separately and equated to the precursor step

• The above rule can be only applied when the given function would be of 0/0 and ∞/∞ form

I would be doing both the processes separately for clarity,not to be confused

Numerator

$\boxed{\begin{minipage}{8 cm} \longmapsto\:\tt\dfrac{d(e^{\sin x}-1)}{dx}\\\\ \longmapsto\:\tt\dfrac{d(e^{\sin x})}{dx}-\dfrac{d(1)}{dx}\\\\ \longmapsto\:\tt e^{\sin x}\qquad\quad---(1)\end{minipage}}$

• Derivative of an exponential function is the very same
• Derivative of a constant is zero

Denominator

$\boxed{\begin{minipage}{7 cm} \tt{ \dfrac{d[log(1 + x)] }{dx} } \\ \\ \longmapsto \: \tt{ \dfrac{1}{x + 1} } - - - - - (2) \end{minipage}}$

From expressions (1) and (2),we get :

$\lim \: \: \: \: \: \: \: \: \: \tt{\dfrac{ {e}^{sin \: x} }{ (\dfrac{1}{x + 1}) } } \\ \tt{x \longrightarrow \: 0}$

Putting x = 0,we get :

$\hookrightarrow \: \tt{ {e}^{sin \: 0} } \\ \\ \huge{ \hookrightarrow \: \tt{0}} \:$

Thus at x = 0,the limit of the above function is zero