Question

Limxtends to infinityx/modulus of x..?

Answer 1

Please see below.

Explanation:

I am interpreting "`x` tends to infinity" as meaning "`x` increases without bound".

`x/absx = {(1,"if",x > 0),(-1,"if",x < 0):}`

Therefore `lim_(xrarroo) x/absx = 1`

Bonus Answer

As `x` tends to minus infinity (as `x` decreases without bound), we have

`lim_(xrarr-oo) x/absx = -1`