Key Concepts and Formulas

• Limit Evaluation using Taylor Series Expansion: For functions that are indeterminate forms (like $\frac{0}{0}$) at a point, Taylor series expansions of elementary functions around that point can be used to evaluate the limit.
  • $\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \cdots$
  • $\sin(u) = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \cdots$
• Indeterminate Form $\frac{0}{0}$: A limit of the form $\frac{0}{0}$ requires further analysis, often involving L'Hôpital's Rule or Taylor series expansion.
• Equating Coefficients of Polynomials: If a polynomial identity holds for all values of a variable, then the coefficients of corresponding powers of the variable on both sides must be equal.

Step-by-Step Solution

Step 1: Simplify the Limit Expression The given limit is: $\lim _{x \rightarrow 1^{+}} \frac{(x-1)(6+\lambda \cos (x-1))+\mu \sin (1-x)}{(x-1)^3}=-1$ To simplify the evaluation, let $t = x-1$. As $x \rightarrow 1^{+}$, we have $t \rightarrow 0^{+}$. Also, $1-x = -t$. Substituting these into the limit expression, we get: $\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda \cos t)+\mu \sin (-t)}{t^3}=-1$ Since $\sin(-t) = -\sin(t)$, the expression becomes: $\lim _{t \rightarrow 0^{+}} \frac{t(6+\lambda \cos t)-\mu \sin t}{t^3}=-1$

Step 2: Apply Taylor Series Expansion We need to evaluate the limit of a fraction as $t \rightarrow 0^{+}$. The denominator $t^3$ approaches 0. For the limit to be a finite non-zero value (-1), the numerator must also approach 0, and the lowest power of $t$ in the numerator must be at least $t^3$. We use the Taylor series expansions of $\cos t$ and $\sin t$ around $t=0$:

• $\cos t = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \cdots$
• $\sin t = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \cdots$

Substitute these expansions into the numerator: $\text{Numerator} = t\left(6+\lambda\left(1 - \frac{t^2}{2!} + \cdots\right)\right) - \mu\left(t - \frac{t^3}{3!} + \cdots\right)$ $\text{Numerator} = t\left(6+\lambda - \frac{\lambda t^2}{2} + \cdots\right) - \mu t + \frac{\mu t^3}{6} - \cdots$ $\text{Numerator} = 6t + \lambda t - \frac{\lambda t^3}{2} + \cdots - \mu t + \frac{\mu t^3}{6} - \cdots$ Group terms by powers of $t$: $\text{Numerator} = (6 + \lambda - \mu)t + \left(-\frac{\lambda}{2} + \frac{\mu}{6}\right)t^3 + \cdots$

Now, substitute this back into the limit expression: $\lim _{t \rightarrow 0^{+}} \frac{(6 + \lambda - \mu)t + \left(-\frac{\lambda}{2} + \frac{\mu}{6}\right)t^3 + \cdots}{t^3}=-1$

Step 3: Ensure the Limit is Finite and Non-Zero For the limit to be a finite non-zero value (-1) as $t \rightarrow 0^{+}$, the terms in the numerator with powers of $t$ less than $t^3$ must be zero. This is because if the coefficient of $t$ (the lowest power) were non-zero, the limit would be $\lim_{t \to 0^+} \frac{Ct}{t^3} = \lim_{t \to 0^+} \frac{C}{t^2}$, which would tend to $\infty$ if $C \neq 0$.

Therefore, the coefficient of $t$ must be zero: $6 + \lambda - \mu = 0 \quad \cdots (\text{i})$ This equation ensures that the numerator starts with a $t^3$ term, making the limit potentially finite.

Step 4: Evaluate the Limit with the Condition from Step 3 With $6 + \lambda - \mu = 0$, the limit expression simplifies to: $\lim _{t \rightarrow 0^{+}} \frac{\left(-\frac{\lambda}{2} + \frac{\mu}{6}\right)t^3 + \text{higher order terms}}{t^3}=-1$ $\lim _{t \rightarrow 0^{+}} \left(-\frac{\lambda}{2} + \frac{\mu}{6}\right) + \text{terms with } t^2 \text{ and higher} = -1$ As $t \rightarrow 0^{+}$, all terms with $t^2$ and higher powers go to zero. Thus, the limit is equal to the coefficient of the $t^3$ term: $-\frac{\lambda}{2} + \frac{\mu}{6} = -1 \quad \cdots (\text{ii})$

Step 5: Solve the System of Linear Equations We now have a system of two linear equations with two variables, $\lambda$ and $\mu$: (i) $\lambda - \mu + 6 = 0 \implies \lambda - \mu = -6$ (ii) $-\frac{\lambda}{2} + \frac{\mu}{6} = -1$

Let's solve this system. Multiply equation (ii) by 6 to clear the denominators: $-3\lambda + \mu = -6 \quad \cdots (\text{ii'})$ Now we have: (i) $\lambda - \mu = -6$ (ii') $-3\lambda + \mu = -6$

Add equation (i) and equation (ii') to eliminate $\mu$: $(\lambda - \mu) + (-3\lambda + \mu) = -6 + (-6)$ $-2\lambda = -12$ $\lambda = 6$ Substitute the value of $\lambda$ into equation (i): $6 - \mu = -6$ $\mu = 6 + 6$ $\mu = 12$

Step 6: Calculate $\lambda + \mu$ We have found $\lambda = 6$ and $\mu = 12$. The question asks for the value of $\lambda + \mu$. $\lambda + \mu = 6 + 12 = 18$

Common Mistakes & Tips

• Incorrect Taylor Series: Ensure accurate recall and application of Taylor series for $\sin(u)$ and $\cos(u)$. A common mistake is using only the first term or misremembering the signs or factorials.
• Forgetting the $t$ in the Numerator: The term $(x-1)$ in the numerator multiplies the entire $(6+\lambda \cos(x-1))$ term. When substituting $t=x-1$, this becomes $t(6+\lambda \cos t)$, not just $6+\lambda \cos t$.
• Equating Coefficients Incorrectly: For the limit to be finite, the coefficient of the lowest power of $t$ in the numerator that is less than the power of $t$ in the denominator must be zero. In this case, the coefficient of $t$ must be zero.

Summary The problem involves evaluating a limit that results in an indeterminate form. By substituting $t=x-1$, the limit is transformed into an expression involving $t$. The Taylor series expansions of $\sin t$ and $\cos t$ are then used to represent the numerator as a polynomial in $t$. For the limit to be a finite non-zero value, the coefficients of the terms in the numerator with powers of $t$ less than the denominator's power ($t^3$) must be zero. This leads to a system of two linear equations for $\lambda$ and $\mu$. Solving this system yields $\lambda=6$ and $\mu=12$, and their sum $\lambda+\mu$ is 18.

The final answer is $\boxed{18}$.