L'Hospital's Rule; Indeterminate Forms. How do I get the answer to this?

Question

Everyway I've tried has failed. Please help

Answer 1

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Answer 2

l'Hospital's rule gives you a mess!

Try this:

Factor x² out of each radical and ignore the constant term in each case since it is of higher order.

You get xsqrt[1+(8/x)]-xsqrt[1+(1/x)]

Use the binomial expansion of the sqrt to get

x[1+(4/x)+terms of higher order]-x[1+(1/2)(1/x)+terms of higher order]=3.5

and this looks correct graphically.

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Answer 3

(sqrt(x^2+8x-5)- sqrt(x^2+1x+6)) * (sqrt(x^2+8x-5)+sqrt(x^2+1x+6))/ (sqrt(x^2+8x-5)+sqrt(x^2+1x+6)) =

(x^2+8x-5 -(x^2+1x+6))/(sqrt(x^2+8x-5)+sqrt(x^2+1x+5)) =

(7x - 11)/(sqrt(x^2+8x-5)+sqrt(x^2+1x+5))

At this point, you are ready to use l'Hopital's rule (though the answer is obvious)

You get

7/(x/sqrt(x^2+8x-5) + x/sqrt(x^2+1x+6))

This has limit 7/(1+1) = 7/2

(If you wish, you can use the squeeze theorem to see that the limit of each term in the denominator is 1.

For the first, as x^2+6x+9 < x^2+8x-5 < x^2+8x+16 for x >7, x/(x+3) < x/sqrt(x^2+8x-5) < x/(x+4), and each of x/(x+3) and x/(x+4) has limit 1; similarly, x^2 < x^2+1x+6 < x^2+2x+1 for x > 5, so 1 < x/sqrt(x^2+1x+6) < x/(x+1), and 1 and x/(x+1) have limits of 1.)