Question

Hi there!

Can I get some help with number 16 and 20 please? Thanks! Show your work :)


Thanks!

Answer 1

16. We could write

`(2x+3)/(x-3)=(2(x-3)+9)/(x-3)=2+\frac9{x-3}`

(basically find the quotient/remainder upon dividing
`2x+3` by
`x-3`). Then as
`x\to3`,
`\frac9{x-3}` diverges to
`-\infty` from the left, and to
`+\infty` from the right. So the limit does not exist.

20. Factorize the numerator:

`(x^2+x-6)/(x-2)=((x-2)(x+3))/(x-2)`

We're considering the limit as
`x\to2`, which also means that it's not the case that
`x=2`. Because of this, we can cancel the factor of
`x-2` in both numerator and denominator:

`\displaystyle\lim_(x\to2)(x^2+x-6)/(x-2)=\lim_(x\to2)x+3`

`x+3` exists for all values of
`x` (i.e. it's continuous on its domain), so the limit is the value of
`x+3` at
`x=2`, so

`\displaystyle\lim_(x\to2)x+3=2+3=5`