Question

Does the limit `lim_(x rarr oo) e^x/x^2 rarr oo` exist? If so evaluate it.

Answer 1

`+oo`

Explanation:

The simple explanation is that the `e^x` function increases much faster than the `x^2` function, so as we reach arbitrarily large positive x, the numerator will become orders of magnitude larger than the denominator.

The slightly more complex method is L'HOPITAL'S RULE, which states that `lim_(x->n)f(x)/g(x) = lim_(x->n)(f'(x))/(g'(x))`. This can be extended to higher order derivatives, such as second order (e.g. `(f''(x))/(g''(x)`, etc.

The derivative of the numerator is simply `e^x`; same for the second derivative, the third, etc. The derivative of `x^2` is `2x`, and the second order derivative is 2.

Using this, we can find...

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

Since we know this limit is simply `+oo` for the second derivative, it is also true for the first derivative and the original function.

Answer 2

`lim_(x rarr oo) e^x/x^2 rarr oo`

Explanation:

If we look at the power series (Maclaurin Series) for `e^x` we have:

`e^x = 1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ...`

Dividing by `x^2`, we get:

`e^x/x^2 = 1/x^2{1 + x + (x^2)/(2!) + (x^3)/(3!) + (x^4)/(4!) + (x^5)/(5!) + ... }`
`\ \ \ \ \ = 1/x^2 + 1/x + 1/(2!) + (x)/(3!) + (x^2)/(4!) + (x^3)/(5!) + ...`

Now, if we take the limit as `x rarr oo`, we have:

`lim_(x rarr oo) e^x/x^2 = lim_(x rarr oo){1/x^2 + 1/x + 1/(2!) + (x)/(3!) + (x^2)/(4!) + ... }`

Clearly the first two terms vanish as `x rarr oo`, leaving

`lim_(x rarr oo) e^x/x^2 = lim_(x rarr oo){ 1/(2!) + {x)/(3!) + (x^2)/(4!) + (x^3)/(5!) + ... }`

And for large `x`, each term increase without bound, and so therefore does the sum of those unbound terms, thus:

`lim_(x rarr oo) e^x/x^2 rarr oo`