Question

use l'hopital's rule to evaluate the limit \[\lim_{x \rightarrow3}\frac{ x-1-\sqrt{x ^{2}-5} }{ x-3 }\]

Answer 1

where are you stuck ?

Answer 2

Remember, L'Hopital's Rule states that if f(a) = g(a) = 0, and that f'(a) and g'(a) exist, and that g'(a) is not equal to 0. Then:\[\lim_{x \rightarrow a}\frac{ f(x) }{ g(x) }=\frac{ f'(a) }{ g'(a) }\]So to apply the rule, we first check if both the top and bottom, i.e f(a) = g(a) = 0, and for us a = 3. Plugging in 3 gives us both the 0 in the numerator and the denominator, hence indeterminate form. So now we take the derivative of both f(x) at a and g(x) at a. f(x) is the numerator and g(x) is the denominator. Let's take the derivative of each:\[f'(x)=1-\frac{ 1 }{ 2 }(x^2-5)^{-1/2}(2x) \]\[g'(x) = 1\]So now we evaluate f'(3)/g'(3) to get the limit.\[\lim_{x \rightarrow 3}\frac{ f(x) }{ g(x) }=\frac{ -0.5 }{ 1 }=-\frac{ 1 }{ 2 }\]
@MATTW20

Answer 3

Thank you, that makes sense

Answer 4

Fan me? lol :D @mattw