Question

please ans limit class 12th​

Answer 1

Question:- To find the (x->$\infty$$lim \: n \times( \frac{f(x + t) + 2f(x + 2t) + .... + nf(x + nt)}{f(x + t) + 4f(x + 4t) + .... {n}^{2}f(x + {n}^{2}t)})$

given that,

$f(x + t) = f(x)$

now, if we analyse

$f(x + 2t) = f(x + t + t)$

putting f(x+t)=f(x)

we get

$f(x + t + t) = f(x + t) = f(x)$

so,

$f(x + 2t) = f(x)$

similarly

$f(x + 2t) = f(x + 3t) = f(x + 4t) = f(x + 5t) = f(x)$

putting this value in given terms we get

$= > n \times \frac{f(x) + 2(fx) + 3(fx) + ...... + nf(x)}{f(x) + 4f(x) + ..... + {n}^{2}fx }$

now, taking out f(x) common from both numerator and denominator we get

$= >n \times \frac{f(x)(1 + 2 + 3 + .... +n) }{f(x)(1 + 4 + 9 + .... + {n}^{2}) }$

cancelling out f(x) from numerator and denominator

we get

=>

$n \times \frac{1 + 2 + 3 + .... + n}{ {1}^{2} + {2}^{2} + {3}^{2} + ... {n}^{2} }$

as we know,

sum of first n-natural number=

$\frac{n(n + 1)}{2}$

sum of square of first n-natural number=

$\frac{ n(n + 1)(2n + 1)}{6}$

putting these value we get

limit

$(x = > \infty ) = \frac{ {n}^{2}(n + 1) \times 6 }{n(n + 1)(2n + 1) \times 2}$

cancelling out (n+1) from numerator. and denominator we get

$= > \frac{ {n} \times 6}{(2n + 1) \times 2}$

now divide the numerator and denominator by n

=>$\frac{3}{2 + \frac{1}{n} }$

now putting n=infinite we get

required limit=$\frac{3}{2 + \frac{1}{ \infty } } = \frac{3}{2 + 0} = \frac{3}{2}$

Hence 3/2 will be the required limit