## Basic limits?

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### Basic limits?

Im studying for a Calculus test I have on monday and the professor gave us a list of problems to do to practice for the test. Heres one that I stumbled upon and havent been able to solve:

limit as x approaches -2 of ((sin(x^2+3x+2)/(x+2))

Please dont solve it or give me the answer, could anyone tell me whats the correct method or theorem to apply? Somehow I just cant get it. Thanks.
LuisRodg

Posts: 12
Joined: Sun Sep 02, 2007 7:19 am

Since x -> -2 and your denominator is (x+2) => you can't divide by zero.
You either need to simplify or use a rule that "deals" with dividing by zero.
Since you have sin(expression) on top, simplifying the expression is not likely (possible).
Evaluating the top also gives us zero (sin(0) = 0). So we have 0/0 situation.
As far as I remember there was a theorem dealing with that kind of stuff...

ventsyv

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Joined: Mon Sep 22, 2003 5:25 pm
Location: MD USA

Yes that would be by applying L'Hopital Rule of type 0/0 however the professor wants us to find it without using HR.

Seriously I have stared at this problem for like 20 minutes trying to device a way to do it but no luck...Any help is appreciated. According to the solutions manual the answer is -1 but that doesnt help me, I want to know how to do it.
LuisRodg

Posts: 12
Joined: Sun Sep 02, 2007 7:19 am

[tex]
lim x-> -2 (sin(x^2 + 3x + 2) / (x+2)) = lim x-> -2 ( d/dx (sin(x^2 + 3x + 2) / (x+2)))
[/tex]

Edit: thats the HR, never mind. Hey anyone knows why is tex not working ?

ventsyv

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Joined: Mon Sep 22, 2003 5:25 pm
Location: MD USA

lim x -> -2 (f/g) = lim x -> -2 (d/dx (f) / g)

Is that even a rule?

LuisRodg

Posts: 12
Joined: Sun Sep 02, 2007 7:19 am

no I just remembered that lim ( f(x) / g(x) ) = lim ( f`(x) / g`(x) )
But now that I think about it thats what the HP is, so ...

Here is another idea:

x^2 + 3x + 2 = (x+2)(x+1)

Are you sure the question is not :

lim x-> -2 ( sin ( (x^2+3x+2)/ (x+2) ) )

ventsyv

Posts: 2810
Joined: Mon Sep 22, 2003 5:25 pm
Location: MD USA

Completely sure. Im looking at it in the book right now.

Thing is that:

x^2 + 3x + 2 = (x+2)(x+1)

is just messing with the argument of the sin function, not much i can do by factoring I would guess?,
LuisRodg

Posts: 12
Joined: Sun Sep 02, 2007 7:19 am

If you know that
lim x->0 sin(x)/x = 1
then
lim x->0 x/sin(x) = 1
and you can prove then that lim x->-2 (x^2+3x+2)/sin(x^2+3x+2) = 1

Now you can multiply your original expression by something whose limit at -2 is 1, and that won't change the result. Cancel out the sin(...) parts and perform the polynomial division.

That would be my plan.

Alvaro
Moderator

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Location: NY, USA

I talked with my professor today and he showed me this solution:

limit as x approaches -2 of ((sin(x^2+3x+2)/(x+2))

As Alvaro said, we need to use the theorem that lim x->x1 (sin(x1)/x1) = 1

limit as x approaches -2 of ((sin(x^2+3x+2)/(x+2))

limit as x approaches -2 of ((sin((x+2)(x+1))/(x+2))

limit as x approaches -2 of [((sin((x+2)(x+1))/(x+2)(x+1)) * (x+1)]

Let z = (x+2)(x+1) , z goes to 0 as x goes to -2

limit as z goes to 0 of (sin(z)/z) * limit as x goes to -2 of (x+1)

1 * ((-2) + 1)

1 * -1

-1

DAMN! hehe

Appreciate all the help you guys have provided, this problem was a bitch.
LuisRodg

Posts: 12
Joined: Sun Sep 02, 2007 7:19 am

Yeah sure is. I'm kind of rusty too.
I was helping some other dude who was trying to do limit of series and find out if it is converging. I told him for one of them it is but then I realized it is not.
I have not done any limits in like 2-3 years, so I'm starting to forget I suppose.

ventsyv

Posts: 2810
Joined: Mon Sep 22, 2003 5:25 pm
Location: MD USA