If lim _{x rightarrow 1^{+}} frac{(x-1)(6+lam | vedclass

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(A) Let $x-1 = h$. As $x \rightarrow 1^{+}$,$h \rightarrow 0^{+}$.Substituting this into the limit,we get $\lim _{h \rightarrow 0} \frac{h(6+\lambda \

Vedclass · Accepted Answer

$18$

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(A) Let $x-1 = h$. As $x \rightarrow 1^{+}$,$h \rightarrow 0^{+}$.Substituting this into the limit,we get $\lim _{h \rightarrow 0} \frac{h(6+\lambda \cos h) - \mu \sin h}{h^3} = -1$.Using the Taylor series expansions $\cos h = 1 - \frac{h^2}{2!} + \frac{h^4}{4!} - \dots$ and $\sin h = h - \frac{h^3}{3!} + \dots$,the expression becomes:$\lim _{h \rightarrow 0} \frac{h(6 + \lambda(1 - \frac{h^2}{2})) - \mu(h - \frac{h^3}{6})}{h^3} = -1$.$\lim _{h \rightarrow 0} \frac{(6 + \lambda - \mu)h + (\frac{\mu}{6} - \frac{\lambda}{2})h^3}{h^3} = -1$.For the limit to exist and equal $-1$,the coefficient of $h$ must be $0$:$6 + \lambda - \mu = 0 \implies \mu - \lambda = 6$.The coefficient of $h^3$ must be $-1$:$\frac{\mu}{6} - \frac{\lambda}{2} = -1 \implies \mu - 3\lambda = -6$.Solving the system: $\mu - \lambda = 6$ and $\mu - 3\lambda = -6$.Subtracting the equations: $2\lambda = 12 \implies \lambda = 6$.Substituting $\lambda = 6$ into $\mu - \lambda = 6$,we get $\mu = 12$.Therefore,$\lambda + \mu = 6 + 12 = 18$.

If $\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$,where $\lambda, \mu \in \mathbb{R}$,then $\lambda+\mu$ is equal to

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