求limit的方法

since sin x = x - (1/3!) x^3 + (1/5!) x^5 + ....

where the kth term : (-1) ^ ( k-1) / (2 k-1) ! (x ^ (2k-1) ) and k from 1 to infinity



then (x^2+6x+7)/sin x =( 1 + 6/x + 7 / x^2 ) / ( 1/ x - 1/6 x + 1/5 x^3 + .....)

when x goes to infinity , the denominator trends to 0

therefore (x^2+6x+7)/sin x trends to infinty.

你要將個條式除以擁有最多指數既x
某d數會變左?/x...x^2....x^3之類
而個x係無限大
咁個數就會冇左
再計落去就計到