Question

i cant answer it pls help​

Answer 1

Let the three consectuive integers be x, x + 2, and x + 4.

Hence, the sum will be given by:

$\implies$x + (x + 2) + (x + 4) = 30 

$\implies$3x + 6 = 30 

$\implies$x = 8

Thus, the three consecutive integers whose sum is 30 are 8, 10 and 12

Answer 2

Showing that the above limit exists we need to perform the following steps:-

$\frac{1}{n + 1} \leqslant integrating \: from \: n \: to \: n + 1 \\ \frac{1}{x} dx \leqslant \frac{1}{n} - \: i)$

Summing the right hand inequality shows that

$summation \frac{1}{k} - log(n + 1) - ii)$

is increasing. Now summing the left hand inequality shows that

$summation \: \frac{1}{k} - log(n) - iii)$

is decreasing.

Since limit n ( infinity) log

$( \frac{n + 1}{n} ) = 0 \: \: 2and \: 3$

shows that given limit exists between 0 and 1.

Learn More:-

Limits formula:- Let y = f(x) as a function of x. If at a point x = a, f(x) takes indeterminate form, then we can consider the values of the function which is very near to a. If these values tend to some definite unique number as x tends to a, then that obtained a unique number is called the limit of f(x) at x = a.