How do I find the limit as x -> 2 of [sqrt(6-x)-2]/[sqrt(3-x)-1]?

Question

PERFECT first answer, but imagine now, if you will, that my big problem with this stems from my having to pretend that I don't know how to do derivatives yet (because it's only the second week of class, and the professor wouldn't like that). So I have to do it some other way...

Answer 1

lim (x->2) [√(6-x)-2] / [√(3-x)-1]
Both numerator and denominator tend to 0, so try L'Hopital's Rule. Differentiate top and bottom to get
lim (x->2) [(1/2)(6-x)^(-1/2).(-1)] / [(1/2)(3-x)^(-1/2).(-1)]
= lim (x->2) [√(3-x) / √(6-x)]
= √1 / √4
= 1/2.
Since this limit exists, the original limit also exists and is 1/2.

Answer 2

lim (x->2) [√(6-x)-2] / [√(3-x)-1]

Both numerator and denominator tend to 0, so try L'Hopital's Rule.

Differentiate top and bottom to get;
lim (x->2) [(1/2)(6-x)^(-1/2).(-1)] / [(1/2)(3-x)^(-1/2).(-1)]

= lim (x->2) [√(3-x) / √(6-x)]

= √1 / √4

= 1/2................ans

Since this limit exists, the original limit also exists and is 1/2.

Answer 3

L=[sqrt(6-x)-2] / [sqrt(3-x)-1];
♦ let z=x-2, so z→0; L=[sqrt(4-z)-2] / [sqrt(1-z)-1];
= 2[sqrt(1-z/4)-1] / [sqrt(1-z)-1];
♦ Let u=sqrt(1-z), u’(z)=(-1/2)/u(z);
for z→0, u(z) =u(1) +u’(1)z +o(z) =1-(1/2)z +o(z);
♦ Let v=sqrt(1-z/4), v’(z)=(-1/8)/v(z);
for z→0, v(z) =v(1) +v’(1)z +o(z) =1-(1/8)z +o(z);
♦♦ L=2(1-(1/8)z +o(z) -1) / (1 -(1/2)z +o(z) -1) = (-1/4) /(-1/2) =1/2;

Answer 4

I don't think there is a way without L'Hospitals rule, i.e. using derivatives. I'm a math tutor.