Question

What is the limit of `(x^2 + x- 30)/|x- 5|` as x approaches infinity?

Answer 1

`lim_(xrarroo)(x^2+x-30)/abs(x-5)=oo`

For positive values of `x`, `abs(x-5)=x-5`. So,

`lim_(xrarroo)(x^2+x-30)/abs(x-5)=lim_(xrarroo)(x^2+x-30)/(x-5)`.

This limit evaluates to the indeterminate form `oo/oo`.

(If you have access to l'Hopital's rule, you could use it, but it is not necessary.)

Use algebra to make the denominator not go to `oo`.

For all positive `x` (which is all we're interested in as `xrarroo`),

`(x^2+x-30)/abs(x-5)=(x^2+x-30)/(x-5)=(x(x+1-30/x))/(x(1-5/x))=(x+1-30/x)/(1-5/x)`

As `x` increases without bound, the numerator of this expression also increases without bound and the denominator approaches `1`.

Therefore:`lim_(xrarroo)(x^2+x-30)/abs(x-5)=lim_(xrarroo)(x+1-30/x)/(1-5/x)=oo`