If `alpha in(0,1)` and `f:R->R` and `lim_(x->oo)f(x)=0,lim_(x->oo)(f(x)-f(alphax))/x=0,` then `lim_(x->oo)f(x)/x=lambda` where `2lambda+7` is

Answer

7

Explanation

`lim_(x to oo) (f(x)-f(alphax))/x=0`
`implies` For any `epsilongt0` there is `deltagt0` such that `|x|lt delta` and `|f(x)-f(alphax)|lt epsilon |x|`
using triangle inequality `|f(x)-f(alpha^(n)x)le|f(x)-f(alpha x)|+|f(alpha^(2)x)|+..+|f(alpha^(n-1)x)-f(alpha^(n)x)|`
`gt epsilon|x|(1+alpha+alpha^(2)+........+alpha^(n-1))=epsilon (1-alpha^(n))/(1-alpha)|x| le(epsilon|x|)/(1-alpha)`
As, `nto oo`
`|f(x)|le (epsilon)/(1-alpha)|x|`
`implieslim_(x to oo) (f(x))/x=0`