Question

The value of
\[ \lim_{n\to\infty} n\int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt \]equals _______ (answer in integer).

Hint:

As \(n\to\infty\), the interval shrinks to the single point \(t=1\), so the expression tends to the value of \(e^{t^2-1}\) at \(t=1\).

Answer 1

Correct Answer: 1

Step 1: Recognise the structure of the limit.
We need
\[ L=\lim_{n\to\infty} n\int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt. \]
The interval of integration is centred at \(t=1\) and has width \(\left(1+\frac{1}{2n}\right)-\left(1-\frac{1}{2n}\right)=\frac{1}{n}.\) As \(n\to\infty\), this width shrinks to \(0\), so the integral is taken over a tiny interval around \(t=1\).

Step 2: Apply the Mean Value Theorem for integrals.
Since \(f(t)=e^{t^2-1}\) is continuous, the Mean Value Theorem for integrals says there exists a point \(c_n\) in the interval \(\left(1-\frac{1}{2n},\,1+\frac{1}{2n}\right)\) such that
\[ \int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt = e^{c_n^2-1}\times\left(\frac{1}{n}\right). \]

Step 3: Multiply by \(n\).
\[ n\int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt = n\times e^{c_n^2-1}\times\frac{1}{n}=e^{c_n^2-1}. \]

Step 4: Take the limit as \(n\to\infty\).
Because \(c_n\) is squeezed between \(1-\frac{1}{2n}\) and \(1+\frac{1}{2n}\), and both of these tend to \(1\) as \(n\to\infty\), we get \(c_n\to1.\) So
\[ \lim_{n\to\infty} e^{c_n^2-1}=e^{1^2-1}=e^0=1. \]

Step 5: Verify with Leibniz differentiation as a cross check.
An equivalent way is to see the limit as a derivative. Write \(g(h)=\int_{1-h}^{1+h} e^{t^2-1}\,dt\), \(h=\frac{1}{2n}\). Then \[ n\int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt = \frac{g(h)}{2h}. \] Since \(g(0)=0\), by L'Hopital's rule and the Leibniz rule for differentiating under the integral sign, \[ \lim_{h\to0}\frac{g(h)}{2h}=\lim_{h\to0}\frac{g'(h)}{2}=\lim_{h\to0}\frac{e^{(1+h)^2-1}+e^{(1-h)^2-1}}{2}=\frac{e^0+e^0}{2}=1. \] Both methods agree.

Final Answer:
The limit evaluates to the value of \(e^{t^2-1}\) at \(t=1\), which is \(1\).\[ \boxed{1} \]