Step 1: Spot that this is an averaging limit. The expression is \[ L=\lim_{n\to\infty} n\int_{1-\frac{1}{2n}}^{1+\frac{1}{2n}} e^{t^2-1}\,dt \] Note that $n=\dfrac{1}{\text{width of interval}}$, since the interval $\left[1-\dfrac{1}{2n},\,1+\dfrac{1}{2n}\right]$ has length $\dfrac1n$. So $n$ times the integral is just the average value of $e^{t^2-1}$ over that shrinking interval.
Step 2: Use a first order expansion near $t=1$. Write $t=1+s$ where $s$ ranges over $\left[-\dfrac{1}{2n},\dfrac{1}{2n}\right]$, a small interval around $0$. Then \[ t^2-1=(1+s)^2-1=2s+s^2 \] For small $s$, $2s+s^2\approx 2s$, so \[ e^{t^2-1}\approx e^{2s}\approx 1+2s \] using the fact that $s\to0$ as $n\to\infty$.
Step 3: Average this approximation over the small interval. \[ n\int_{-\frac{1}{2n}}^{\frac{1}{2n}} (1+2s)\,ds = n\left[s+s^2\right]_{-\frac{1}{2n}}^{\frac{1}{2n}} \] Evaluating, the $s$ term gives $n\left(\dfrac{1}{2n}-\left(-\dfrac{1}{2n}\right)\right)=n\cdot\dfrac1n=1$, and the $s^2$ term gives $n\left(\dfrac{1}{4n^2}-\dfrac{1}{4n^2}\right)=0$ since the square is symmetric and cancels out.
Step 4: Take the limit. So the leading behaviour gives exactly $1$, and all the higher order correction terms from the approximation vanish faster than $\dfrac1n$ as $n\to\infty$, so they contribute nothing in the limit.
Step 5: Confirm using the direct evaluation at the centre. More simply, as the interval shrinks to the single point $t=1$, the average value of the continuous function $e^{t^2-1}$ over the interval converges to its value at the centre point, which is \[ e^{1^2-1}=e^0=1 \] This matches the expansion method. \[ \boxed{1} \]
