In order to compute this limit, we need to rationalize the numerator, and in order to rationalize the numerator we need to multiply the numerator and denominator by the conjugate of √(x^2 + x) - √(x^2 - x), which is √(x^2 + x) + √(x^2 - x).
So, multiplying the numerator and denominator by √(x^2 + x) + √(x^2 - x) yields:
TA is correct at all times ;) when you plug in x=zero you get [0/0] which is an indeterminate. Practice L'Hop rule: lim x => 0 (2x) / -sinx ===>still [0/0] repeat L'Hop lim x=> zero (2) / -cosx ===> subsequently 2/(-1) = -1 answer = -2 also, you must know lim x=>zero of sinx / x = x /sinx each ==> zero if you happen to have no idea L'Hopital's rule, then multiply the numerator and denominator by means of the conjugate: (x^2) / (cosx -1) multiply top/backside with the aid of (cosx +1) x^2 (cosx + 1) / (cos^2(x) - 1) ====> use Pyth id sin^2x+cos^2x=1 =x^2(cosx + 1) / -sin^2(x) = - ( x/sinx )( x/sinx ) (cosx + 1) every sinx / x limit goes to one as x goes to zero therefore = - (cos x + 1) = -(cos0 + 1) = -(1+1) = -2 :)
See the cited source below, and, once you get there, choose the selection called "show steps" to see out l'hospital's rule had been used several times, along with some very hairy algebra, to obtain the solution.
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In order to compute this limit, we need to rationalize the numerator, and in order to rationalize the numerator we need to multiply the numerator and denominator by the conjugate of √(x^2 + x) - √(x^2 - x), which is √(x^2 + x) + √(x^2 - x).
So, multiplying the numerator and denominator by √(x^2 + x) + √(x^2 - x) yields:
lim (x-->infinity) [√(x^2 + x) - √(x^2 - x)]/3
= lim (x-->infinity) {[√(x^2 + x) + √(x^2 - x)][√(x^2 + x) - √(x^2 - x)]}/{3[√(x^2 + x) + √(x^2 - x)]}
= lim (x-->infinity) [(x^2 + x) - (x^2 - x)]/{3[√(x^2 + x) + √(x^2 - x)]}
(The above step was done via difference of squares.)
= lim (x-->infinity) 2x/{3[√(x^2 + x) + √(x^2 - x)]}
= 2/3 * lim (x-->infinity) x/[√(x^2 + x) + √(x^2 - x)], by pulling out 2/3.
Now, notice that for x > 0:
(a) √(x^2 + x) = √[x^2 * (1 + 1/x)] = x√(1 + 1/x)
(b) √(x^2 - x) = √[x^2 * (1 - 1/x)] = x√(1 - 1/x).
So, the limit now becomes:
2/3 * lim (x-->infinity) x/[x√(1 + 1/x) + x√(1 - 1/x)].
Dividing the numerator and numerator by x yields the required limit to be:
2/3 * lim (x-->infinity) 1/[√(1 + 1/x) + √(1 - 1/x)] = (2/3)[1/(1 + 1)] = 1/3.
I hope this helps!
TA is correct at all times ;) when you plug in x=zero you get [0/0] which is an indeterminate. Practice L'Hop rule: lim x => 0 (2x) / -sinx ===>still [0/0] repeat L'Hop lim x=> zero (2) / -cosx ===> subsequently 2/(-1) = -1 answer = -2 also, you must know lim x=>zero of sinx / x = x /sinx each ==> zero if you happen to have no idea L'Hopital's rule, then multiply the numerator and denominator by means of the conjugate: (x^2) / (cosx -1) multiply top/backside with the aid of (cosx +1) x^2 (cosx + 1) / (cos^2(x) - 1) ====> use Pyth id sin^2x+cos^2x=1 =x^2(cosx + 1) / -sin^2(x) = - ( x/sinx )( x/sinx ) (cosx + 1) every sinx / x limit goes to one as x goes to zero therefore = - (cos x + 1) = -(cos0 + 1) = -(1+1) = -2 :)
1/3
See the cited source below, and, once you get there, choose the selection called "show steps" to see out l'hospital's rule had been used several times, along with some very hairy algebra, to obtain the solution.
That is a NASTY problem.
Ciao!
alice
Rationalize the numerator.