lim----> -2 (x^2-4)/(x^2+3x+2)?

Answer 1

numerator = (x-2)(x+2)

denominator = (x+1)(x+2)
ratio = (x-2)/(x+1)
putting x = -2 we get limit = (-2-2)/(-2+1) = -4/-1 or 4

Answer 2

lim x -->-2 (x² - 4)/(x² + 3x + 2)

When we substitute we see that it yields the indeterminate form 0/0. There are two methods in solving this. One is the convenient L'Hopital's Rule. If it yields the indeterminate form 0/0, then you can get the derivatives of both the numerator and denominator. Thus,
= lim x-> -2 (2x)/(2x + 3)

Now we can substitute:
= (-4)/(-4 + 3)
= -4/(-1)
Thus the limit is
= 4

but if you still do not know how to get the derivative, we can simply factor both the numerator and the denominator and cancel the factor that is common to them. thus,
= lim x->-2 (x + 2)(x - 2)/(x + 2)(x + 1)

Thus,
= lim x->-2 (x - 2)/(x + 1)

We can now substitute
= (-2 - 2)/(-2 + 1)

Thus,
= -4/-1

And the limit still is
= 4

^_^

Answer 3

Well since it's algebraic the limit is equal to(with a bit of factorisation)
lim (x+2)(x-2) / (x+2)(x+1)
x->-2

Now you can see that the (x+2) cancels off in the numerator to that in the denominator.
So even though a zero is involved in the strictest sense, we may as well simplify to the fullest before evaluating the limit.
i.e. the limit becomes
lim (x-2) / (x+1)
x->-2

Now just substitute the -2 for x since we won't get any nasty 0/0 s,etc.
then,
(-2-2)/(-2+1) = -4/-1 = 4

Now I'll give you a never fail method should you ever run into problems with factorisation.
The result is L'Hopital's Rule and implies that for two functions f and g, f(a)=0 AND g(a)=0
lim f(x) / g(x) = lim f'(x) / g'(x)
x->a x->a

where the dash notation implies the derivative. Similarly as long as the functional fraction is undefined it is equal to any n th derivative ratio. a can be any value, even complex, zero or infinity.

For your case,
lim (x^2 - 4) / (x^2 + 3x + 2) = lim 2x / 2x+3 = 2(-2)/(-1)=4
x->-2 x->-2

as required.
Note that for y = ax^n
dy/dx = nax^(n-1)

Hope this helps!

Answer 4

lim x----> -2 (x^2-4)/(x^2+3x+2)
= lim x--> -2 [(x - 2)(x + 2)]/[(x + 1)(x + 2)] by factorising numerator and denominator
= lim x--> -2 (x - 2)/(x + 1) [as (x + 2) is a common factor]
= (-2 - 2)/(-2 + 1) by substitution
= -4/-1
= 4

Answer 5

numerator = (x-2)(x+2)

denominator = (x+1)(x+2)
as x--> -2 then (x+2)--->0 i.e (x+2) not equal to 0
ratio = (x-2)/(x+1)
putting x = -2 we get limit = (-2-2)/(-2+1) = -4/-1 = 4