Question

lim x tends to 0 sin(e^x-3 -1)/log(2-x)​

Answer 1

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Question :

$\bullet\sf{lim_{x \to 0} \: \: \dfrac{sin(e^ (x-3)-1)}{log(2-x)}}$

Answer :

Let, first see that it is in the form $\dfrac{0}{0}$ or not.

$\implies\displaystyle\tt{ lim_{ x \to 0 } \: \: \: \dfrac{sin(e^(0-3)-1)}{log(2-0)}}$

$\implies\displaystyle\tt{ lim_{ x \to 0} \: \: \: \dfrac{sin(e^-3-1)}{log2}}$

$\implies\displaystyle\tt{lim_{x \to 0} \: \: \: \dfrac{sin(\dfrac{1}{e^3}-1)}{log2}}$

$\implies\displaystyle\tt{lim_{ x \to 0} \: \: \: \dfrac{sin(1-e^3)}{e^3log2}}$

$\\ \implies{\boxed{\underline{\bf{we \: get, lim_{x \to 0} \: \: \: \dfrac{sin(1-e^3)}{e^3 log2} \: \: in \: \dfrac{0}{0} form}}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Method (2)

• By using L - Hospital Rule to remove $\dfrac{0}{0}$ form:

$\implies\displaystyle\tt{lim_{x \to 0} \: \: \dfrac{dy}{dx}\dfrac{sin(e^x-3)-1)}{log(2-x)}}$

$\implies\displaystyle\tt{lim_{ x \to 0} \: \: \dfrac{cos(e^(x-3)-1) \times e^x \times 1}{\dfrac{1}{(2-x)} }}$

$\implies\displaystyle\tt{lim_{ x \to 0} \: \: \dfrac{ cos(e^(0-3)-1) \times e^0 \times 1 }{\dfrac{1}{2-0}}}$

$\implies\displaystyle\tt{lim_{x \to 0} \: \: 2cos(e^-3)-1}$

$\boxed{\underline{\bf{we \: get, lim_(x \to 0) \: \: 2cos(e^-3)-1 \: \: from \: L -hospital \: Rule}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬