Math Help - annoying limit

1. annoying limit

Hey
i've just been trying to work out this limit
its been hours and my tutor rekons its just - infinite
i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

here it is
(lim as x approaches -3 from the right)

lim 5x^2/ x^2 + 2x - 3
x->-3+

Thanks in advance

2. Originally Posted by Daniiel
Hey
i've just been trying to work out this limit
its been hours and my tutor rekons its just - infinite
i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

here it is
(lim as x approaches -3 from the right)

lim 5x^2/ x^2 + 2x - 3
x->-3+

Thanks in advance
Dear Daniiel,

By looking at the graph of this function you will understand it more clearly.

3. Originally Posted by Daniiel
Hey
i've just been trying to work out this limit
its been hours and my tutor rekons its just - infinite
i've tried dividing all by x cubed, factorizing and trying to cancel, multiplying but the conjugant and messing around with it

here it is
(lim as x approaches -3 from the right)

lim 5x^2/ x^2 + 2x - 3
x->-3+

Thanks in advance
Try using long division:

$\frac{5x^2}{x^2 + 2x - 3} = 5 + \frac{15 - 10x}{x^2 + 2x - 3}$.

Now use partial fractions on the second term:

$\frac{A}{x + 3} + \frac{B}{x - 1} = \frac{15 - 10x}{x^2 + 2x - 3}$

$\frac{A(x - 1) + B(x + 3)}{(x + 3)(x - 1)} = \frac{15 - 10x}{x^2 + 2x - 3}$.

Since the denominators are equal, so must be the numerators:

$A(x - 1) + B(x + 3) = 15 - 10x$

$Ax - A + Bx + 3B = 15 - 10x$

$3B - A + (A + B)x = 15 - 10x$

So $3B - A = 15$ and $A + B = -10$

Solving these equations simultaneously gives

$A = -\frac{45}{4}$ and $B = \frac{5}{4}$.

Therefore

$5 + \frac{15 - 10x}{x^2 + 2x - 3} = 5 - \frac{45}{4(x + 3)} + \frac{5}{4(x - 1)}$.

Does this help?

4. I did somthing similar to that but when you do lim x>-3 of the product you still get an undefined faction
but i might be wrong

does the graph show that as it approaches -3 that it isnt infinte?

5. The important term is the $-\frac{45}{4(x + 3)} = -\frac{45}{4}\left(\frac{1}{x + 3}\right)$.

This is a basic hyperbola. You should know enough about hyperbolae of the form $\frac{1}{x - a}$ to be able to say what happens as you approach $a$ from each side.

6. What you should have seen immediately is that the numerator, $5x^2$ does not go to 0 at x= -3. Since the denominator does, that alone is enough to tell you that the limit does not exist and all your " dividing all by x cubed, factorizing and trying to cancel, multiplying by the conjugant and messing around with it" is useless. Saying it goes to $-\infty$ is just saying it does not exist for a particular reason- and you can see that by looking at the graph or by evaluating the function at, say, -2.9.