Question

`lim_(x->oo)((x^2+4x+1)/(x^2+2x+2))^(10x)=` ?

Answer 1

`Lim_(xrarroo)((x^2+4x+1)/(x^2+2x+2))^(10x)=e^20`

Explanation:

We know that: `Lim_(xrarroo)(1+1/x)^x=e`

`Lim_(xrarroo)((x^2+4x+1)/(x^2+2x+2))^(10x)`

`Lim_(xrarroo)((x^2+4x+1+2x-2x+1-1)/(x^2+2x+2))^(10x)`

`Lim_(xrarroo)((x^2+2x+2)/(x^2+2x+2)+(2x-1)/(x^2+2x+2))^(10x)`

`Lim_(xrarroo)(1+1/((x^2+2x+2)/(2x-1)))^(10x)`

`Lim_(xrarroo)(1+1/((x^2+2x+2)/(2x-1)))^((10x)((x^2+2x+2)/(2x-1))((2x-1)/(x^2+2x+2))`

Looks very messy but I like it this way. You can substitute to make it more clear but as you can see now we have the same formula as in the beginning.

`Lim_(xrarroo)[(1+1/((x^2+2x+2)/(2x-1)))^((x^2+2x+2)/(2x-1))]^((10x)((2x-1)/(x^2+2x+2))`

We know that: `Lim_(xrarroo)[(1+1/((x^2+2x+2)/(2x-1)))^((x^2+2x+2)/(2x-1))]=e`

And now we can put limit to the exponent (superscript)

`e^(Lim_(xrarroo)(10x)((2x-1)/(x^2+2x+2)))=?`

Let's evaluate limit and then finish it

`Lim_(xrarroo)(10x)((2x-1)/(x^2+2x+2))`

`Lim_(xrarroo)((20x^2-10x)/(x^2+2x+2))`

`Lim_(xrarroo)(cancel(x^2)(20-10/x))/(cancel(x^2)(1+2/x+2/x^2))`

`(20-10/oo)/(1+2/oo+2/oo^2)=(20-0)/(1+0+0)=20/1=20`

The answer: `Lim_(xrarroo)((x^2+4x+1)/(x^2+2x+2))^(10x)=e^20`

Answer 2

`e^20`

Explanation:

First

`((x^2+4x+1)/(x^2+2x+2))^(10x) approx ((1+4/x)/(1+2/x))^(10x)` for big `x`

Second, making the transformation

`(1+4/x)/(1+2/x)=1+1/y rArr x = 2(y-1)` and then

`((1+4/x)/(1+2/x))^(10x)=(1+1/y)^(20(y-1)) =( (1+1/y)^y)^20(1+1/y)^-20`

Third

`lim_(x->oo)((1+4/x)/(1+2/x))^(10x)=lim_(y->oo)( (1+1/y)^y)^20(1+1/y)^-20 = (lim_(y->oo) (1+1/y)^y)^20lim_(y->oo)(1+1/y)^-20`

but

`(lim_(y->oo) (1+1/y)^y)^20 = e^20` and `lim_(y->oo)(1+1/y)^-20=1` so finally

`lim_(x->oo)((x^2+4x+1)/(x^2+2x+2))^(10x)=e^20`