Question

How do you evaluate the limit `((x+5)^2-25)/x` as x approaches `0`?

Answer 1

`10`.

Explanation:

Reqd. Limit `=lim_(xrarr0){(x+5)^2-25}/x`

`=lim_(xrarr0){x^2+10x+25-25)/x`

`=lim_(xrarr0){cancelx(x+10)}/cancelx`

`=lim_(xrarr0)(x+10)`

`=10`.