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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.


 \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\\

So the series x_n/n converges to the value c.