Let's break down each part of the problem and solve them step-by-step.
### Part (a)
Evaluate the limit:
[tex]\[
\lim_{n \to \infty} \left(\frac{1}{n}\right)^{\frac{1}{n}}
\][/tex]
1. Start by taking the natural logarithm of the expression:
[tex]\[
\ln\left[\left(\frac{1}{n}\right)^{\frac{1}{n}}\right] = \frac{1}{n} \ln\left(\frac{1}{n}\right)
\][/tex]
2. Simplify the inside logarithm:
[tex]\[
\frac{1}{n} \ln\left(\frac{1}{n}\right) = \frac{1}{n} \cdot (-\ln(n)) = -\frac{\ln(n)}{n}
\][/tex]
3. Evaluate the limit of the simplified expression as [tex]\( n \to \infty \)[/tex]:
[tex]\[
\lim_{n \to \infty} -\frac{\ln(n)}{n}
\][/tex]
It's known that [tex]\(\frac{\ln(n)}{n} \)[/tex] approaches 0 as [tex]\( n \to \infty \)[/tex]:
[tex]\[
\lim_{n \to \infty} -\frac{\ln(n)}{n} = 0
\][/tex]
4. Therefore, the limit of the original expression becomes:
[tex]\[
e^0 = 1
\][/tex]
So, the limit is:
[tex]\[
1
\][/tex]
### Part (b)
Evaluate the integral:
[tex]\[
\int_0^p e^{-x} \, dx
\][/tex]
1. Compute the integral of [tex]\( e^{-x} \)[/tex]:
[tex]\[
\int e^{-x} \, dx = -e^{-x} + C
\][/tex]
2. Evaluate the integral from 0 to [tex]\( p \)[/tex]:
[tex]\[
\left[ -e^{-x} \right]_{0}^{p} = -e^{-p} - (-e^{0})
\][/tex]
3. Simplify the expression:
[tex]\[
-e^{-p} + 1 = 1 - e^{-p}
\][/tex]
So, the value of the integral is:
[tex]\[
1 - e^{-p}
\][/tex]
### Part (c)
Evaluate the limit:
[tex]\[
\lim_{n \to \infty} \left( \sqrt{n^2 - 1} - (n - 1) \right)
\][/tex]
1. Rewrite the expression inside the limit:
[tex]\[
\sqrt{n^2 - 1} - (n - 1) = \sqrt{n^2 - 1} - n + 1
\][/tex]
2. Factor out [tex]\( n \)[/tex] from the square root using the expansion [tex]\(\sqrt{n^2 - 1} \approx n \sqrt{1 - \frac{1}{n^2}}\)[/tex]:
[tex]\[
\sqrt{n^2 - 1} \approx n \sqrt{1 - \frac{1}{n^2}} \approx n \left(1 - \frac{1}{2n^2}\right)
\][/tex]
3. Substitute this approximation back into the original expression:
[tex]\[
n \left(1 - \frac{1}{2n^2}\right) - n + 1 = n - \frac{1}{2n} - n + 1 = 1 - \frac{1}{2n}
\][/tex]
4. As [tex]\( n \to \infty \)[/tex]:
[tex]\[
\frac{1}{2n} \approaches 0
\][/tex]
So, the limit is:
[tex]\[
1
\][/tex]
### Summary
(a) [tex]\[
\lim_{n \to \infty} \left(\frac{1}{n}\right)^{\frac{1}{n}} = 1
\][/tex]
(b) [tex]\[
\int_0^p e^{-x} \, dx = 1 - e^{-p}
\][/tex]
(c) [tex]\[
\lim_{n \to \infty} \left( \sqrt{n^2 - 1} - (n - 1) \right) = 1
\][/tex]
