Question

\lim _{x\to 0}\left(\frac{\sqrt{1+3x+x^2}-1}{\arcsin \left(2x\right)}\right)

Answer 1

Step-by-step explanation:

$\displaystyle \lim_{x \to 0} \left(\dfrac{\sqrt{x^2 + 3x + 1} - 1}{\arcsin 2x} \right)$

Note that as $x \rightarrow 0$, the ratio becomes undefined. Using L'Hopital's Rule, where

$\displaystyle \lim_{x \to c} \dfrac{f(x)}{g(x)} = \lim_{x \to c} \dfrac{f'(x)}{g'(x)}$

where f'(x) and g'(x) are the derivatives of the functions f(x) and g(x), respectively. Note that

$f(x) = \sqrt{x^2 + 3x + 1} \:\:\text{and}\:\: g(x) = \arcsin 2x$

$f'(x) = \dfrac{2x + 3}{2(\sqrt{x^2 + 3x + 1})}$

$g'(x) = \dfrac{2}{\sqrt{1 - 4x^2}}$

Therefore,

$\displaystyle \lim_{x \to 0} \dfrac{f'(x)}{g'(x)} = \lim_{x \to 0} \dfrac{2x + 3}{2(\sqrt{x^2 + 3x + 1})} \times \left(\dfrac{\sqrt{1 - 4x^2}}{2} \right)$

or

$\displaystyle \lim_{x \to 0} \left(\dfrac{\sqrt{x^2 + 3x + 1} - 1}{\arcsin 2x} \right) = \dfrac{3}{4}$