Question

How do you find the limit of `(x^2+3)/(x^2+4)` as `x->-oo`?

Answer 1

`lim_(x->-oo)(x^2+3)/(x^2+4) = 1`

Explanation:

If we look at the graph of `y=(x^2+3)/(x^2+4)` we can see that it is clear that the limit exists, and is approximately `1`

graph{(x^2+3)/(x^2+4) [-10, 10, -2, 2]}

Now, As `x->-oo` then `x^2->oo` ,but `1/x^2->0`

So, it would be better if we could replace `x^2` with `1/x^2`, or `x^-2`

`lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)((x^2+3))/((x^2+4)) * x^-2/x^-2`

`:. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(x^-2(x^2+3))/(x^-2(x^2+4))`

`:. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo) (x^-2x^2+3x^-2) / (x^-2x^2+4x^-2)`

`:. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo) (1+3x^-2) / (1+4x^-2)`

And, using `x^-2->0` as `x->-oo` we have;

`lim_(x->-oo)(x^2+3)/(x^2+4) = (1+0) / (1+0) = 1`

Which is completely consistent with the above graph.

Answer 2

`lim_(x->-oo)(x^2+3)/(x^2+4) = 1`

Explanation:

We can also use L'Hospital's Rule as the limit is of an indeterminate form `oo/oo`

So By L'Hospital's Rule;

`lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)((x^2+3)')/((x^2+4)')`
`:. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(2x)/(2x)`
`:. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(1)`
`:. lim_(x->-oo)(x^2+3)/(x^2+4) = 1`