finding the limit at x approaches 2 on (x^2+x-6) / (x^2-4)?

Question

since its undefined when x = 2, can you apply l'hopitals rule and the limit is 5/4?
also for x approaches 0 for e^x-1-x/ sin[x], is it eventually 0 after you apply the rule? thanks people

Answer 1

The proper condition of using L'Hospital's rule is if it's in the form [0/0], not if it's undefined. That is, if you find the limit as x approaches 0 of 1/x, you cannot use L'Hospital's rule because it's in the form [1/0] after plugging in the value, not [0/0].

In your case though, you can apply L'Hospital's rule if you wanted, but it is much simpler not to.

x^2 + x - 6 factors into (x - 2)(x+3), and
x^2 - 4 factors into (x -2) (x + 2).
As you can see, they have a common factor which can cancel out. Thus

Lim (x --> 2, [x^2 + x - 6]/[x^2 - 4]) =
Lim (x --> 2, [(x - 2)(x+3)]/[(x -2) (x + 2)]) = (at this point you cancel)
Lim (x --> 2, (x+3)/(x+2)) = (2+3)/(2+2) = 5/4

For the second limit, you definitely need to apply L'Hospital's rule, since you get the form 0/0.

Lim (x --> 0, (e^x - 1 - x)/sin(x)) =
Lim (x --> 0, (e^x - 1)/cos(x)) = (e^0 - 1)/cos(0) = (1 - 1)/1 = 0/1 = 0.

Answer 2

since its undefined when x = 2, can you apply l'hopitals rule and the limit is 5/4? yes


for x approaches 0 for e^x-1-x/ sin[x]

e^(x-1-x)/ sin[x] is undefined as x approaches 0

Answer 3

Factor it first:

(x^2+x-6)/(x^2-4) = ((x-2)(x+3))/((x-2)(x+2)) = (x+3)/(x+2)

this is a removable discontinuity. the limit is then 5/4

Answer 4

1) Yes, you can apply la règle de l'Hôpital because the limit is 0 for both numerator and denominator.

2) Yes also. lim(x-->0) e^x-1-x/ sin[x] = 0/0, so
lim(x-->0) e^x-1-x/ sin[x] =
lim(x-->0) (e^x-1)/cosx =
0/1.

Answer 5

5/4 is right on the first one.

The second one isn't 0 it's -1.
e^(x)-1-x/sinx becomes this after lhopitals rule:

e^x-1-1/cosx
1-1-1= -1

Not exactly sure of how the second problem is supposed be set up. I assumed this:

e^(x)-1-(x/sinX)

Answer 6

evaluate the function f(x) = [(x^2?9)/(x?3)]. what might you assume the cut back of f(x) as x procedures 3 to be? right here we've, f(x) = [(x^2 ? 9) / (x ? 3)] f(x) = [(x + 3)(x ? 3) / (x ? 3)] = (x + 3) Now, limx? 3 f(x) = limx? 3 (x + 3)] = 6 < ...................> answer