# Thread: [SOLVED] Some help with limits

1. ## [SOLVED] Some help with limits

I have to evaluate the following limits:

a) lim x -> 7
(square root(x+2)) -3 / (x-7

I was unable to solve this

b) lim x -> 0
1/x - 1/(x^2 + x)

Somehow simplified it to 0 / x+1... this makes no sense.

c) lim x -> -infinity
(square root(4x^2 + 1)) / (3x-5)

I somehow got denominator 0 and was unable to solve.

What I did was take the x^2 out of the square root on top and got:

x^2 square root(4+1/x^2) / 3/x - 5/x^2

I am pretty sure the denominators go to zero and and this point I'm lost.

Please help. I can do this and have done similar problems before, just need a little kick start.

2. Originally Posted by thekrown
I have to evaluate the following limits:

a) lim x -> 7
(square root(x+2)) -3 / (x-7

I was unable to solve this

rationalize the numerator ... multiply by $\textcolor{red}{\frac{\sqrt{x+2} + 3}{\sqrt{x+2} + 3}}$

b) lim x -> 0
1/x - 1/(x^2 + x)

Somehow simplified it to 0 / x+1... this makes no sense.

simplifies to $\textcolor{red}{\frac{1}{x+1}}$

c) lim x -> -infinity
(square root(4x^2 + 1)) / (3x-5)

first, note that the limit will be less than 0. divide the numerator and denominator by x ...

$\textcolor{red}{\frac{\sqrt{4 + \frac{1}{x^2}}}{3 - \frac{5}{x}}}$
...

3. a) Okay gotcha, I can do this.

b) I will redo this, I must of made a repetitive mistake. I will check my work again.

c) I thought in this type of limit we would have to divided by the greatest power of the numerator, in this case x^2. How come just x?

4. Originally Posted by thekrown
a) Okay gotcha, I can do this.

b) I will redo this, I must of made a repetitive mistake. I will check my work again.

c) I thought in this type of limit we would have to divided by the greatest power of the numerator, in this case x^2. How come just x?
if x goes under the radical, what does it become?

5. Let me see if I understand this here... see my teacher uses a robotic-teaching method where he says "just do this" and it works. So i haven't learned the proper reasoning to this type of limit.

If what I think you're saying is, the numerator will divide by the greatest power so our square root(4x^2+1) is divided by x^2 to get square root(4+1/x).

Then for the denominator, we use the square root of x^2 and divide the numerator by that?

Is that why it's just x?

6. Originally Posted by thekrown
Let me see if I understand this here... see my teacher uses a robotic-teaching method where he says "just do this" and it works. So i haven't learned the proper reasoning to this type of limit.

If what I think you're saying is, the numerator will divide by the greatest power so our square root(4x^2+1) is divided by x^2 to get square root(4+1/x).

Then for the denominator, we use the square root of x^2 and divide the numerator by that?

Is that why it's just x?
the full monty ...

$\frac{\sqrt{4x^2+1}}{|x|} = \frac{\sqrt{4x^2+1}}{\sqrt{x^2}} = \sqrt{\frac{4x^2+1}{x^2}}
$

now understand that you're really dividing by $|x|$ since $\sqrt{x^2} = |x|$

no effect on the numerator, since it is positive for all x.

for the denominator, however ...

$\frac{3x-5}{|x|} = -3 - \frac{5}{|x|}$ since $x < 0$