# Finding the limit as x-> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

1. Jan 10, 2013

### Lo.Lee.Ta.

Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

1. lim[√(5 + 5x^2)]/(5 +7x)
x→∞

2. Alright, I thought I would have first find the largest exponent of x in the denominator.

In this case, the largest exponent is x^1.
The next step is to divide every term by x^1.

Since I cannot divide something in a square root by x, I thought I COULD multiply it by
√(1/x^2). That's the same thing as dividing by x.

So, this is what I have:

[√(1/x^2)*√(5 + 5x^2)]/(5/x + 7x/x)

= [ √(5/(x^2) + 5x^2/x^2) / ((5/x) + 7) ] * (1/x^2)/(1/x)

= √((5/x^2) + 5) / ((5/x) + 7)

Then I thought if you substitute infinity for x here, then the (5/x^2) and the 5/x both equal 0.

So, it's √((5/0) + 5) / (0 + 7)

= √(5)/7

.....This is not the right answer... =_=
Could you find my mistakes?
Thank you so much!

2. Jan 10, 2013

### symbolipoint

Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

I was able to transform the expression to:
EDIT: The TEX was just too tedious correct after making it wrong, so I am corrected my mistake in plain text:
(sqrt((5/x^2)+5))/((5/x)+7)

Upon letting x increase without end, I found the same limit as you did,
$\frac{\sqrt{5}}{7}$

3. Jan 10, 2013

### mtayab1994

Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

You're answer is correct:

$$\lim_{x\rightarrow+\infty}\frac{\sqrt{5x^{2}+5}}{7x+5}=\lim_{x\rightarrow+\infty}\sqrt{\frac{5x^{2}+5}{(7x+5)^{2}}}=\lim_{x \rightarrow +\infty} \sqrt{\frac{5x^2+5}{49x^2+70x+25}}=\lim_{x \to +\infty}\sqrt{\frac{5x^2}{49x^2}}=\frac{\sqrt{5}}{7}$$

Last edited by a moderator: Jan 10, 2013
4. Jan 10, 2013

### symbolipoint

Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

Worked solution, because it may be somewhat different from LoLeeTa and mtayab1994.
(Attached)

#### Attached Files:

• ###### lim_sqrt_ratio.JPG
File size:
32.1 KB
Views:
135
5. Jan 10, 2013

### Lo.Lee.Ta.

Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

Thanks so much symbolipoint and mtayab1994!

You guys are awesome! :)

There were 2 aspects to the computer hw problem.
One part was as x→∞, and the other was as x→-∞

So it actually counted the √(5)/7 right for x→∞

BUT for x→-∞, the answer had to be -√(5)/7!

Thanks! :D

6. Jan 10, 2013

### Staff: Mentor

Re: Finding the limit as x--> infinity: [sqrt(5 + 5x^2)]/(5 + 7x)

That's because $\sqrt{x^2}$ = -x if x < 0. Keep in mind that the square root of a nonnegative number is nonnegative, by definition.

In essence you are factoring x2 out of the terms insided the radical, and bringing it out of the radical as -x.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook