If $\lim_{x\to1} \frac{(x-1)(6 + \lambda \cos(x-1)) + \mu \sin(1-x)}{(x-1)^3} = -1$ where $\lambda, \mu \in \mathbb{R}$ then $\lambda + \mu$ is equal to:

Step 1: Substitute $t = x-1$

To simplify the limit expression, we introduce a substitution. Let $t = x-1$. As $x$ approaches $1$, $t$ approaches $0$. This transformation makes the limit easier to evaluate using standard Maclaurin series expansions.

Step 2: Rewrite the limit in terms of $t$

After substituting $t = x-1$, the expression becomes $\frac{t(6 + \lambda \cos(t)) + \mu \sin(-t)}{t^3}$. We also use the property that $\sin(-t) = -\sin(t)$.

Step 3: Apply Maclaurin series expansions

We use the Maclaurin series expansions for $\cos(t) = 1 - \frac{t^2}{2!} + \frac{t^4}{4!} - \dots$ and $\sin(t) = t - \frac{t^3}{3!} + \frac{t^5}{5!} - \dots$. We only need terms up to $t^3$ in the numerator to match the denominator.

Step 4: Simplify the expression

Expand the terms in the numerator and group them by powers of $t$. This helps us identify the coefficients of $t$ and $t^3$.

Step 5: Equate coefficients to solve for $\lambda$ and $\mu$

For the limit to be finite and non-zero, the coefficient of $t$ in the numerator must be zero. This gives us $6 + \lambda - \mu = 0$. Then, the limit becomes the coefficient of $t^3$, so $\frac{\mu}{6} - \frac{\lambda}{2} = -1$.

Step 6: Solve the system of equations

Substitute the first equation into the second: $(\lambda + 6) - 3\lambda = -6$. This simplifies to $-2\lambda = -12$, so $\lambda = 6$. Then, $\mu = 6 + 6 = 12$.

Step 7: Calculate $\lambda + \mu$

Finally, we calculate the sum of $\lambda$ and $\mu$ using the values we found.

More from this chapter

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