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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 |
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Answer» Correct Answer - 7 `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` |
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