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

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Answer 1 · 2017-06-27T19:57:29

$e^{4}$ $e^4$

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

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 $e^4$