can't solve a limite! help!?

Question

lim
x->4

3 - (squareroot) (x^2-7)
-------------------------------
x-4

the squareroot includes x^2-7 and the dot line means division.i know that is 0/0 but i can't solve it
please help!

Answer 1

3 - √(x² - 7)
---------------- =
x - 4

[3 - √(x² - 7)] [3 + √(x² - 7)]
---------------- ------------------- =
[x - 4] [3 + √(x² - 7)]

9 - (x² - 7)
---------------- ---------- =
[x - 4] [3 + √(x² - 7)]

16 - x²
---------------- ---------- =
[x - 4] [3 + √(x² - 7)]

(4 - x)(4 + x)
---------------- ---------- =
[x - 4] [3 + √(x² - 7)]

-(x + 4)
-------------------
[3 + √(x² - 7)]

So lim x → 4 {(3 - √(x² - 7))/(x - 4)}

= lim x → 4 {(-(x + 4)/[3 + √(x² - 7)]}

= -8/6

= -4/3

Answer 2

The first step is to plug-in x = 4 into the limit. As noted when x = 4,
sqrt(x^2-7) = 3 and 3-3 / 4-4 = 0 / 0. This is a good indication that you should be using L'Hôpital's rule.

Essentially this rule states that if you have a limit of the quotient of two functions,
e.g.
lim x->n [f(x) / g(x)]

Where f and g both approach 0 or infinity
e.g.
lim x->n f(x) = 0 and lim x->n g(x) = 0
OR
lim x->n f(x) = ∞ and lim x->n g(x) = ∞

Then the limit of the quotient of the derivatives of these functions is equal to the limit of the functions themselves.
e.g
lim x->n [f(x) / g(x)] = lim x->n [f'(x) / g'(x)]

In your case f(x) = 3-sqrt(x^2-7) and g(x) = x-4
Since lim x->4 f(x) = 0 and the lim x->4 g(x) = 0 L'Hôpital's rule will apply.

f'(x) = - x / (sqrt(x^2-7) [Using the chain rule]
g'(x) will simply be 1.

Taking the limit of f'(x) / g'(x) we now get
lim x->4 [f'(x) / g'(x)]
= lim x->4 [-x / sqrt(x^2-7)]
= -4 / sqrt(9)
= -4 / 3

Which by L'Hôpital's rule is equal to
lim x->4 [3 - sqrt(x^2-7) / (x-4)].

For more information of L'Hôpital's rule, consult wikipedia: http://en.wikipedia.org/wiki/L%27hopitals_rule

Answer 3

I'm not sure if you've heard of L'Hopitals rule, or if thats too complicated for what you are doing. It is very helpful though: it tells you that when you have 0/0 (or some other cases), you can differentiate the top and bottom.

The top differentiated is -0.5(x^2-7)^-0.5 * 2x, and the bottom differentiated is 1.

Thus, we want the limit as x->4 of (-0.5(x^2-7)^-0.5 * 2x)/1. Substitute x = 4 and you get -4/3.

Answer 4

multiply both the top and bottom by (3+Sqrt[x^2-7])
so we get:
(9-(x^2-7))/[(x-4)(3+Sqrt[x^2-7])]=
(4+x)(4-x)/[(x-4)(3+Sqrt[x^2-7])]=
-(4+x)(x-4)/[(x-4)(3+Sqrt[x^2-7])]=
-(4+x)/(3+Sqrt[x^2-7])
-8/6=
-4/3

Answer 5

Yeah, I would also use L'hopitals rule.