Limits, Continuity & Differentiability Question - None

Question

Paragraph for Questions 630 and 631
Let $f: R \to R$ is a function defined by $f(x) = \begin{cases} 1, & \text{if } x = 1 \ e^{(x^{10}-1)} + (x-1)^2 \sin\left(\dfrac{1}{x-1}\right), & \text{if } x \neq 1 \end{cases}$

If $\lim_{x \to \infty} \left( x \left( \sum_{k=1}^{100} f\left(1 + \dfrac{k}{x}\right) - 100 \right) \right) = \lambda$, then the value of $\dfrac{\lambda}{100}$ is:

Accepted Answer

Step 1: Analyze f(x) for x ≠ 1

For x close to 1, let x = 1 + h where h is small.

f(1+h) = e^((1+h)^10 - 1) + h²sin(1/h)

Since |h²sin(1/h)| ≤ h² → 0 as h → 0, this term vanishes in the limit.

Step 2: Expand e^((1+h)^10 - 1) using Taylor series

(1+h)^10 = 1 + 10h + 45h² + 120h³ + ... (binomial expansion)

(1+h)^10 - 1 = 10h + 45h² + 120h³ + ...

e^(10h + 45h² + 120h³ + ...) = 1 + (10h + 45h² + ...) + (10h + 45h² + ...)²/2 + ...

= 1 + 10h + 45h² + 50h² + O(h³) = 1 + 10h + 95h² + O(h³)

Step 3: Set up the sum as a Riemann sum

∑_{k=1}^{100} f(1 + k/x) = ∑_{k=1}^{100} [1 + 10(k/x) + 95(k/x)² + O(1/x³)]

= 100 + (10/x)∑_{k=1}^{100} k + (95/x²)∑_{k=1}^{100} k² + O(1/x²)

Step 4: Use sum formulas

∑_{k=1}^{100} k = 100(101)/2 = 5050

∑_{k=1}^{100} k² = 100(101)(201)/6 = 338350

Step 5: Calculate the limit

∑_{k=1}^{100} f(1 + k/x) - 100 = (10/x)(5050) + (95/x²)(338350) + O(1/x²)

x[∑_{k=1}^{100} f(1 + k/x) - 100] = 10(5050) + O(1/x) = 50500 + O(1/x)

lim_{x→∞} x[∑_{k=1}^{100} f(1 + k/x) - 100] = 50500

Step 6: Find λ/100

λ = 50500

λ/100 = 50500/100 = 505

∴ Answer: C

Question:

Step-by-Step Solution

Master Limits, Continuity & Differentiability with Mathbee