Question

How do you find the limit of `(2-e^x)/(2+3e^x)` as x approaches infinity?

Answer 1

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

Explanation:

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

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

Now, As `x->oo` then `e^x->oo` ,but `e^-x->0`

So, it would be better if we could replace `e^x` with `e^-x`

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

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

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

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

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

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

Which is completely consistent with the above graph.