explain the questions. each step plz.
Attachments:
nainitha9797:
it is not opening
Answers
Answered by
0
1. Let f(a) = e^a at x = a and let "a" be rational.
In the neighbour hood of a⁻, let there be an irrational number b = a⁻
f(b) = e^(1-b)
As x (and b) approaches a⁻, its value is e^(1-a⁻)
For the function to be continuous at the point "a",
e^a = e^(1-a⁻) a = 1 - a⁻
as a⁻ approaches, "a", 2a = 1 a = 1/2
The function is continuous only at 1/2. At other values of x in (0,1), the function has different values for a⁻, a and a⁺. SO it is not continuous.
2)
![\lim_{n \to \infty} \frac{x_{n+1}}{n+1} = \lim_{n \to \infty} \frac{c+x_{n}}{n+1} = \lim_{n \to \infty} \frac{c+c+x_{n-1}}{n+1}\\ \\ = \lim_{n \to \infty} \frac{c+c+c+x_{n-2}}{n+1} = \lim_{n \to \infty} \frac{nc+x_{1}}{n+1}\\ \\= \lim_{n \to \infty} [ \frac{c}{1+\frac{1}{n}}+\frac{x_1}{n+1} ]=\lim_{n \to \infty} [ \frac{c}{1+0} +\frac{x_1}{n+1} ] = c + 0 = c\\ \lim_{n \to \infty} \frac{x_{n+1}}{n+1} = \lim_{n \to \infty} \frac{c+x_{n}}{n+1} = \lim_{n \to \infty} \frac{c+c+x_{n-1}}{n+1}\\ \\ = \lim_{n \to \infty} \frac{c+c+c+x_{n-2}}{n+1} = \lim_{n \to \infty} \frac{nc+x_{1}}{n+1}\\ \\= \lim_{n \to \infty} [ \frac{c}{1+\frac{1}{n}}+\frac{x_1}{n+1} ]=\lim_{n \to \infty} [ \frac{c}{1+0} +\frac{x_1}{n+1} ] = c + 0 = c\\](https://tex.z-dn.net/?f=+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bx_%7Bn%2B1%7D%7D%7Bn%2B1%7D+%3D++%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bc%2Bx_%7Bn%7D%7D%7Bn%2B1%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bc%2Bc%2Bx_%7Bn-1%7D%7D%7Bn%2B1%7D%5C%5C+%5C%5C+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bc%2Bc%2Bc%2Bx_%7Bn-2%7D%7D%7Bn%2B1%7D+%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5Cfrac%7Bnc%2Bx_%7B1%7D%7D%7Bn%2B1%7D%5C%5C+%5C%5C%3D+%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5B+%5Cfrac%7Bc%7D%7B1%2B%5Cfrac%7B1%7D%7Bn%7D%7D%2B%5Cfrac%7Bx_1%7D%7Bn%2B1%7D+%5D%3D%5Clim_%7Bn+%5Cto+%5Cinfty%7D+%5B+%5Cfrac%7Bc%7D%7B1%2B0%7D+%2B%5Cfrac%7Bx_1%7D%7Bn%2B1%7D+%5D+%3D+c+%2B+0+%3D+c%5C%5C)
So the series x_n/n converges to the value c.
In the neighbour hood of a⁻, let there be an irrational number b = a⁻
f(b) = e^(1-b)
As x (and b) approaches a⁻, its value is e^(1-a⁻)
For the function to be continuous at the point "a",
e^a = e^(1-a⁻) a = 1 - a⁻
as a⁻ approaches, "a", 2a = 1 a = 1/2
The function is continuous only at 1/2. At other values of x in (0,1), the function has different values for a⁻, a and a⁺. SO it is not continuous.
2)
So the series x_n/n converges to the value c.
Similar questions