# find the limit of (x+1)/(3x-2)find the limit of lim x->-infinity (x+1)/(3x-2)

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Find the limit as x decreases without bound of `f(x)=(x+1)/(3x-2)` :

`lim_(x->-oo)(x+1)/(3x-2)`

If we try direct substitution we get the form `(-oo)/(-oo)`

Apply L'hopital's rule: `lim (f(x))/(g(x))=lim(f'(x))/(g'(x))` :

`lim_(x->-oo)(x+1)/(3x-2)=lim_(x->-oo)1/3=1/3`

**Sources:**

We need to factor out the x terms from the numerator and the denominator, but first do a transformation from negative to positive infinity.

Let `u=-x`

Then `u->infty` as `x->-infty` .

`lim_{x->-infty}{x+1}/{3x-2}`

`=lim_{u->infty}{-u+1}/{-3u-2}` factor out the u

`=lim_{u->infty}{u(-1+1/u)}/{u(-3-2/u)}`

`=lim_{u->infty}{-1+1/u}/{-3-2/u}` now take limits

`={-1}/-3`

`=1/3`

**The limit is `1/3` **