Question

Let 𝑓∶ℝ→ℝ be a function defined by
𝑓(𝑥)=𝑥² sin(𝑥-1) + 𝑥𝑒^(𝑥−1) , 𝑥∈ℝ . 
Then, 
\(lim_{𝑛→∞} 𝑛 (𝑓 (1+\frac{1}{n} )+𝑓 (1+\frac{2}{𝑛} )+ ⋯ +𝑓(1+\frac{10}{𝑛} )−10)\)
equals ________

Answer 1

Correct Answer: 165

To find the limit \(lim_{n→∞} n \left(f \left(1+\frac{1}{n} \right)+f \left(1+\frac{2}{n} \right)+ ⋯ +f(1+\frac{10}{n} )−10\right)\), we first consider the function: 
\(f(x) = x^2 \sin(x-1) + xe^{(x-1)}.\)

We approximate \(f(1+\frac{k}{n})\) around \(x=1\) using Taylor series expansion:
For \(x^2 \sin(x-1)\) around \(x=1\): 
\(x^2 \approx 1+\frac{2k}{n}\),
\(\sin(x-1) \approx (x-1) = \frac{k}{n}\),
so, \(x^2 \sin(x-1) \approx (1+\frac{2k}{n}) \cdot \frac{k}{n} \approx \frac{k}{n} + \frac{2k^2}{n^2}.\)
For \(xe^{(x-1)}\) around \(x=1\):
\(xe^{(x-1)} \approx 1+\frac{k}{n} + \left(1+\frac{k}{n}\right)\frac{k}{n} \approx 1+\frac{k}{n}+\frac{k}{n}+\frac{k^2}{n^2}.\)
Thus, \(xe^{(x-1)} \approx 1 + 2\frac{k}{n} + \frac{k^2}{n^2}.\)
Combining both parts:
\(f(1+\frac{k}{n}) \approx 1 + 3\frac{k}{n} + \frac{k^2}{n^2}.\)

The expression inside the limit becomes:
\(\sum_{k=1}^{10} \left(1 + 3\frac{k}{n} + \frac{k^2}{n^2}\right)\).
\(\sum_{k=1}^{10} 1 = 10,\)
\(\sum_{k=1}^{10} 3\frac{k}{n} = \frac{3}{n}\sum_{k=1}^{10} k = \frac{3}{n}\cdot55 = \frac{165}{n},\)
\(\sum_{k=1}^{10} \frac{k^2}{n^2} = \frac{1}{n^2} \cdot \sum_{k=1}^{10} k^2 = \frac{1}{n^2} \cdot 385.\)

The full expression is:
\(n\left(10 + \frac{165}{n} + \frac{385}{n^2} - 10\right) = 165 + \frac{385}{n}.\)
As \(n \to ∞\), \(\frac{385}{n} \to 0\), so the limit is 165.

This result falls within the specified range of 165,165.