How to find infinite limits of a function without plugging in values but instead alegbrically?

1 Answer | Add Yours

embizze's profile pic

Posted on

(1) Infinite limits -- a function is said to have an infinite limit at x=c if the function grows without bound as `x -> c` .(Or decreases without bound.)

The limit does not exist -- we say that the limit is `+- oo` , but this is just a description of how the limit fails to exist.

To determine if a rational function has an infinite limit, first divide out any common factors of the numerator and denominator. Then any factor of the denominator that has a value of zero at x=c causes the function to have a vertical asymptote at x=c. The limit of the function at x=c will be positive or negative infinity. (To determine whether it is positive or negative consider the sign of the numerator and denominator as `x->c^+,x->c^-` .

Ex `lim_(x->-2)(x^2+2x-8)/(x^2-4)`

`=lim_(x->-2)((x+4)(x-2))/((x+2)(x-2))`

`=lim_(x->-2)(x+4)/(x+2)`

Since the numerator is nonzero at x=-2 and the denominator is zero, the function has a vertical asymptote at x=-2; the function increases without bound as `x->-2^+` and the function decreases without bound as `x->-2^-` .

(2) Limits at infinity -- Obviously you cannot substitute `+-oo` for x, so one method is to substitute increasingly larger (smaller) values for x.

Polynomials will have infinite limits at infinity.

Exponential functions will have infinite limits in one direction, and finite limits in the other direction. Ex `lim_(x->oo)2^(-x)=0` and `lim_(x->oo)2^(-x)+k=k` .

Rational functions -- Use `lim_(x->oo)c/(x^r)=0` to find a possible finite limit.

Ex: `lim_(x->oo)(2x-1)/(x+1)` We divide the numerator and denominator by the highest power of x in the expression -- in this case x:

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

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

`=(2-0)/(1+0)=2`

There are many other specialized techniques, e.g. when a function has two horizontal asymptotes, etc...

We’ve answered 331,200 questions. We can answer yours, too.

Ask a question