Question

Let $f\left(x\right)$ be a function such that ${lim}_{x\to 0}\frac{f\left(x\right)}{x}=1$. If ${lim}_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{{\left\{f\right(x\left)\right\}}^3}=1$, then $|a+b|=$

Answer 1

${lim}_{x\to 0}\frac{f\left(x\right)}{x}=1$ As $x\to 0f\left(x\right)$ shauld $\to 0$ for

$\frac{f\left(x\right)}{x}$ to be solvable.

Now let $L={lim}_{x\to 0}\frac{x(1+acosx)-bsinx}{\left(f\right(x){)}^3}$

it is of $\frac{0}{0}$ type,have applying l'opiter,

$L={lim}_{x\to 0}\frac{x(-asinx)+1+acosx-bcosx}{3\left(f\right(x\left){)}^2{f}^{\prime }\right(x)}$

As denominator $\to 0$, Numerator also $\to 0$

$\Rightarrow 1+a-b=0$

$f^{\prime }\left(x\right)$ when $x\to 0$ $={lim}_{x\to 0}\frac{f(o+h)-f\left(0\right)}{h}={lim}_{x\to 0}\frac{f\left(h\right)}{h}=1$

Again applying l'opital,

$L={lim}_{x\to 0}\frac{-asinx-axcosx-asinx+bsinx}{6f\left(x\right)}$

Again, $L={lim}_{x\to 0}\frac{-acosx-acosx+axsinx-acosx+bcos}{6}$

As , $L=1\Rightarrow -a-a-a+b=6\Rightarrow b-3a=6$

As $b=a+1$ (From (1)) $\Rightarrow a+1-3a=6\Rightarrow 2a=-5$

$\Rightarrow a=-5/2,b=-3/2\to |a+b|=|-4|=4$