_x1^+((x-1)(6+(x-1))+(1-x))/((x-1)^3)=-1, where ,R.

Question

$\displaystyle\lim_{x	o1^+}\frac{(x-1)(6+\lambda\cos(x-1))+\mu\sin(1-x)}{(x-1)^3}=-1$, where $\lambda,\mu\in\mathbb{R}$. Then $\lambda+\mu$ is equal to

Accepted Answer

Let $t=x-1\to0^+$. The expression becomes
$$\lim_{t\to0^+}\frac{6t+\lambda t\cos t-\mu\sin t}{t^3}$$

Expanding $\cos t=1-\dfrac{t^2}{2}+\cdots$ and $\sin t=t-\dfrac{t^3}{6}+\cdots$:
$$6t+\lambda t\left(1-\frac{t^2}2\right)-\mu\left(t-\frac{t^3}6\right)=t(6+\lambda-\mu)+t^3\left(-\frac\lambda2+\frac\mu6\right)+\cdots$$

For the limit to be finite, the coefficient of $t$ must vanish: $\lambda-\mu+6=0$ ...(1). Matching the $t^3$ term to the limit value $-1$: $-\dfrac\lambda2+\dfrac\mu6=-1$ ...(2).

Solving (1) and (2): from (1), $\lambda=\mu-6$; substituting into (2): $-\dfrac{\mu-6}{2}+\dfrac\mu6=-1\implies-3(\mu-6)+\mu=-6\implies-2\mu=-24\implies\mu=12,\ \lambda=6$.

$$\lambda+\mu=18$$

Solutions · 1