Question

When `x->2`, `(x^15-2^15)/(x-2)` tends to `__`?

Answer 1

As `xrarr2`, `((x^15-2^15))/((x-2))rarr15xx2^14`

Explanation:

The function `((x^n-a^n))/((x-a))` (if `n` is a natural number) is an interesting function.

Note that `(x^n-a^n)` can be factorized as follows:

`(x^n-a^n)=x^n-ax^((n-1))+ax^((n-1))-a^2x^((n-2))+a^2x^((n-2))-a^3x^((n-3))+...............+x^2a^((n-2))-xa^((n-1))+xa^((n-1))-a^n`

= `x^((n-1))(x-a)+ax^((n-2))(x-a)+a^2x^((n-3))(x-a)+.......+xa^((n-2))(x-a)+a^((n-1))(x-a)`

= `(x-a)(x^((n-1))+ax^((n-2))+a^2x^((n-3))+....+xa^((n-2))+a^((n-1)))`

Hence `(x^15-2^15)=(x-2)(x^14+2^1x^13+2^2x^12+......+2^13x+2^14)`

and `(x^15-2^15)/(x-2)=x^14+2^1x^13+2^2x^12+......+2^13x+2^14`

and as `xrarr2`, `(x^15-2^15)/(x-2)rarr2^14xx15=15xx2^14`