Question

${lim}_{x\to 0}$ $\frac{ax+xcosx}{bsinx}$

Answer 1

${lim}_{x\to 0}\frac{ax+xcosx}{bsinx}\Rightarrow {lim}_{x\to 0}\frac{x(a+cosx)}{bsinx}$

$\Rightarrow$ on using Hospital rule, we get

$\Rightarrow {lim}_{x\to 0}\frac{\frac{d}{dx}\left[\right(a+cosx\left)\right]}{\frac{d}{dx}\left[bsinx\right]}={lim}_{x\to 0}\frac{a+cosx-xsinx}{bcosx}$

again using L Hospital rule we get

$\Rightarrow {lim}_{x\to 0}\frac{-sinx-sinx-xcosx}{-bsinx}$

again using L'Hospital rule

${lim}_{x\to 0}\frac{-2cosx-cosx+xsinx}{-bcosx}$

$\Rightarrow {lim}_{x\to 0}\frac{-3cosx+xsinx}{-bcosx}$

$\Rightarrow {lim}_{x\to 0}+\frac{3}{b}-\frac{xtanx}{b}$

using limit we get

$\Rightarrow \frac{3}{b}$ to = $\frac{3}{b}$