# Find the imit of 2x + 1 - sqrt(4x^2 + 5) as x--> infinity

1. Jan 29, 2008

### projection

1. The problem statement, all variables and given/known data

$$\lim_{x \rightarrow \infty}$$$$2x+1-$$$$\sqrt{4x^2+5}$$

3. The attempt at a solution

i am wondering if this method that i used is correct. i get the correct answer but i ahaven't see it in the text book or on the net. am i doing something that shouldn't be done?

using the limit laws, i take out 1.

$$\lim_{x \rightarrow \infty}$$ 1 + $$\lim_{x \rightarrow \infty$$$$2x-\sqrt{4x^2+5}$$

then i rationalize.
.
$$\lim_{x \rightarrow \infty$$ 1 + $$\lim_{x \rightarrow \infty} \frac{4x^2-4x^2-5}{2x+\sqrt{4x^2+5}}$$

1 + $$\lim_{x \rightarrow \infty}$$$$\frac{-5}{2x+x\sqrt{4+\frac{5}{x^2}}}$$

1 + $$\lim_{x \rightarrow \infty} \frac{\frac{-5}{x}}{\frac{2x}{x}+\frac{x}{x}\sqrt{4x+\frac{5}{x^2}}}$$

1 + $$\frac{0}{4}$$

1+0=1

i get the right answer when i do it for $$\lim_{x \rightarrow \infty$$$$\sqrt{9x^4-3x^2+1} - 3x^2+5$$ and these the only two i tried. is this method just lucky for these two or can i keep using it?

also. i seem to be getting the right limit of 1 when x approches postive infinity for $$\lim_{x \rightarrow \infty$$$$\sqrt{9x^2+6x-5} - 3x$$ but i get -1 when approches negative infinity and mathematica keeps telling me its +infinity. can someone do provide me with a soloution?

Last edited: Jan 30, 2008
2. Jan 29, 2008

### rocomath

I have no idea what your problem says, plz fix your latex!!!

infinity = \infty

lim as x approaches a = \lim_{x \rightarrow a}

3. Jan 30, 2008

### projection

better?

4. Jan 30, 2008

### buzzmath

try pulling the x^2 out of the square root and then so how that affects the terms inside the square root as x goes to inf.

5. Jan 30, 2008

### projection

i tried to solve it, but i keep on ending up with a zero at the bottom. does this mean that what i did is wrong or the limit is infinity whenever you end up with a zero in the denominator?
THE INFINTY IS NEGATIVE. sorry i dont know to syntex of a negative.

$$\lim_{x \rightarrow- \infty$$$$\sqrt{9x^2+6x-5} - 3x$$

$$\lim_{x \rightarrow- \infty$$$$\sqrt{9x^2+6x-5} - 3x$$ . $$\frac{\sqrt{9x^2+6x-5}+3x}{\sqrt{9x^2+6x-5}+3x}$$

$$\lim_{x \rightarrow- \infty$$ $$\frac{6x-5}{\sqrt{9x^2+6x-5}+3x}$$

$$\lim_{x \rightarrow- \infty$$ $$\frac{6x-5}{|x|\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+3x}$$

$$\lim_{x \rightarrow- \infty$$ $$\frac{6-\frac{5}{x}}{\frac{|x|}{x}\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+\frac{3x}{x}}$$

$$\frac{6}{-1\sqrt{9}+3}$$

$$\frac{6}{0}$$

so does this mean the what i did is wrong, or does this indicate the limit is infinity like mathematica keep telling me and what i did is the correct steps?

Last edited: Jan 30, 2008
6. Jan 30, 2008

### CompuChip

Why is $$|x| / x = -1$$ if x goes to + infinity? What is |x| if x becomes very large?
And my Mathematica doesn't tell me it's infinity.

7. Jan 30, 2008

### projection

^^^ oh i forgot to do it there but its going to negative infinity.

i will fix it.

i am getting the correct answer for all positive inf questions. just neg ones i am not getting.

8. Jan 30, 2008

### projection

$$\lim_{x \rightarrow- \infty$$$$\sqrt{9x^2+6x-5} - 3x$$

can someone show me the proper soloution. i tried to find some explanation on the web but no help. i keep getting a zero in the denominator when i use this method.