Question

Lim ((x^2+2x+2) / (x^2+3))^x ?

lim ((x^2+2x+2) / (x^2+3))^x

Answer 1

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

Explanation:

The problem is to evaluate the following limit, which we call `L`

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

To deal with the exponent, take the natural logarithm of both sides

`lnabsL=lim_(x->oo)lnabs(((x^2+2x+2)/(x^2+3))^x)`

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

Note that

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

(by using the ratio of the highest degree terms)

and therefore

`lim_(x->oo)lnabs((x^2+2x+2)/(x^2+3))=ln1=0`

so

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

has the form `oo*0`

Let's rearrange it like so:

`rArrlim_(x->oo)lnabs((x^2+2x+2)/(x^2+3))/(1/x)`

Which is now `0/0`, so we can apply L'Hôpital's Rule.

Taking the derivative of both the numerator and denominator gives:

`rArrlim_(x->oo)((x^2+3)/(x^2+2x+2)*((2x+2)(x^2+3)-(x^2+2x+2)(2x))/(x^2+3)^2)/(-1/x^2)`

What a mess. Let's see if we can simplify it!

`rArrlim_(x->oo)(((2x^3+6x+2x^2+6)-(2x^3+4x^2+4x))/((x^2+2x+3)(x^2+3)))/(-1/x^2)`

`rArrlim_(x->oo)((-2x^2-2x+6)/(x^4+2x^3+6x^2+6x+9))/(-1/x^2)`

`rArrlim_(x->oo)(-x^2(-2x^2-2x+6))/(x^4+2x^3+6x^2+6x+9)`

`rArrlim_(x->oo)(color(red)(2x^4)+2x^3-6x^2)/(color(red)(x^4)+2x^3+6x^2+6x+9)=color(red)2`

Now, we have a limit of a quotient of two polynomials.

It's easy to evaluate this limit by looking at the ratio of the coefficients of the highest degree terms.

(we would arrive at the same conclusion by using L'Hôpital's Rule successively 4 more times)

Now recall that we needed to take the logarithm of the original limit, so we now have:

`lnabsL=2`

and therefore

`color(red)(L=e^2)`