Question

Limx→∞ ((x^2+5x+3)/x^2+x+3))^x=? Thanks.

Answer 1

`e^4`

Explanation:

`lim_(xrarroo) ((x^2+5x+3)/(x^2+x+3))^x`

Step 1
Has indeterminate form `1^oo`. Use exponential and logarithmic functions, with l'Hospital's Rule to find the limit.

Step 2
`((x^2+5x+3)/(x^2+x+3))^x = e^(xln((x^2+5x+3)/(x^2+x+3))`

Step 3
The limit of the exponent is:

`lim_(xrarroo)(xln((x^2+5x+3)/(x^2+x+3)))` which has indeterminate form `0*oo`,

Step 4
but we can rewrite it as

`lim_(xrarroo)(ln((x^2+5x+3)/(x^2+x+3)))/(1/x)` which has form `0/0`.

Step 5
Apply l'Hospital:

`= lim_(xrarroo)((x^2+x+3)/(x^2+5x+3)*(-4(x^2+3))/(x^2+x+3)^2)/(-1/x^2)`

Step 6 `" "` Simplify

`= lim_(xrarroo)(x^2/(x^2+5x+3)*(4(x^2+3))/(x^2+x+3))`

Step 7 Evaluate the limit

`= 4`

Step 8 Use continuity of the exponential function to finish.

`lim_(xrarroo)((x^2+5x+3)/(x^2+x+3))^x = lim_(xrarroo) e^(xln((x^2+5x+3)/(x^2+x+3))`

`= e^(lim_(xrarroo) xln((x^2+5x+3)/(x^2+x+3)))`

`= e^4`

So the limit we seek is `e^4`