Question

What is `lim_(xrarroo) (e^(2x)sin(1/x))/x^2`?

Answer 1

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

Explanation:

Let `y= (e^(2x)sin(1/x))/x^2`

`lny=ln((e^(2x)sin(1/x))/x^2)`

`lny=lne^(2x)+ln(sin(1/x))-lnx^2`

`lny=2xlne+ln(sin(1/x))-2lnx`

`lny=2x+ln(sin(1/x))-2lnx`

`lim_(x->oo) [lny=2x+ln(sin(1/x))-2lnx]`

`lim_(x->oo)lny = lim_(x->oo)[2x+ln(sin(1/x))-2lnx]`

`lim_(x->oo)lny=oo`

`e^lny=e^oo`

`y=oo`

Answer 2

`lim_(xrarroo)(e^(2x)sin(1/x))/x^2 = oo`. Please see the explanation section below.

Explanation:

`lim_(xrarroo)(e^(2x)sin(1/x))/x^2`

Note that: `(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)`

Now, as `xrarroo`, the first ratio increases without bound, while the second goes to `1`.

`lim_(xrarroo)(e^(2x)sin(1/x))/x^2 = lim_(xrarroo)e^(2x)/x^3 * lim_(xrarroo)sin(1/x)/(1/x)`

`= oo`

Further Explanation

Here is the reasoning that led to the solution above.

`lim_(xrarroo)(e^(2x)sin(1/x))/x^2` has initial form `(oo*0)/oo`.

This is an indeterminate form, but we cannot apply l'Hospital's Rule to this form.

We could rewrite it as `(e^(2x))/(x^2/sin(1/x))` to get the form `oo/oo` to which we could apply l'Hospital. However, I don't particularly want to take the derivative of that denominator.

Recall that `lim_(thetararr0)sintheta/theta = 1`.

So that `lim_(xrarroo)sin(1/x)/(1/x) = 1`.

This is what motivates the rewriting used above.

`(e^(2x)sin(1/x))/x^2 = e^(2x)/x^3 * sin(1/x)/(1/x)`.

As `x` increases without bound, `e^x` goes to infinity much faster that `x^3` (faster than any power of `x`).
So, `e^(2x) = (e^x)^2` blows up even faster.

If you do not have this fact available, use l'Hospital's rule to get

`lim_(xrarroo)e^(2x)/x^3 = lim_(xrarroo)(2e^(2x))/(3x^2)`

`= lim_(xrarroo)(8e^(2x))/(6) = oo`