Question

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

Answer 1

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

Explanation:

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

graph{(e^x-x^2)/(e^x+x) [-5, 15, -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)(e^x-x^2)/(e^x+x) = lim_(x->oo)((e^x-x^2)/(e^x+x)) * e^-x/e^-x`

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

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

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

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

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

Which is completely consistent with the above graph.

Answer 2

1

Explanation:

Use iteratively L'Hospital's rule, until the indeterminate form `oo/oo`

disappears..

`lim x to oo ((e^x-x^2)/(e^x+x))`

`=lim x to oo (((e^x-x^2)')/((e^x+1)'))`

`=lim x to oo (((e^x-2)')/((e^x)'))`

`lim x to oo (1-2e^(-x))`

`=(1-0)=1`