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

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In summary, you are trying to solve a limit problem using the limit laws but you are getting a zero in the denominator whenever you use the method.
  • #1
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Homework Statement



[tex]\lim_{x \rightarrow \infty}[/tex][tex] 2x+1-[/tex][tex]\sqrt{4x^2+5}[/tex]



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 textbook or on the net. am i doing something that shouldn't be done?

using the limit laws, i take out 1.

[tex]\lim_{x \rightarrow \infty}[/tex] 1 + [tex]\lim_{x \rightarrow \infty [/tex][tex]2x-\sqrt{4x^2+5}[/tex]

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

1 + [tex]\lim_{x \rightarrow \infty} [/tex][tex]\frac{-5}{2x+x\sqrt{4+\frac{5}{x^2}}}[/tex]

1 + [tex]\lim_{x \rightarrow \infty} \frac{\frac{-5}{x}}{\frac{2x}{x}+\frac{x}{x}\sqrt{4x+\frac{5}{x^2}}}[/tex]

1 + [tex]\frac{0}{4}[/tex]

1+0=1

i get the right answer when i do it for [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^4-3x^2+1} - 3x^2+5[/tex] 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 [tex]\lim_{x \rightarrow \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex] but i get -1 when approches negative infinity and mathematica keeps telling me its +infinity. can someone do provide me with a soloution?
 
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  • #2
I have no idea what your problem says, please fix your latex!

infinity = \infty

lim as x approaches a = \lim_{x \rightarrow a}
 
  • #3
better?
 
  • #4
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
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 don't know to syntex of a negative.

[tex]\lim_{x \rightarrow- \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex]


[tex]\lim_{x \rightarrow- \infty [/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex] . [tex]\frac{\sqrt{9x^2+6x-5}+3x}{\sqrt{9x^2+6x-5}+3x}[/tex]

[tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6x-5}{\sqrt{9x^2+6x-5}+3x}[/tex]

[tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6x-5}{|x|\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+3x}[/tex]

[tex]\lim_{x \rightarrow- \infty [/tex] [tex]\frac{6-\frac{5}{x}}{\frac{|x|}{x}\sqrt{9+\frac{6}{x}-\frac{6}{x^2}}+\frac{3x}{x}}[/tex]


[tex]\frac{6}{-1\sqrt{9}+3}[/tex]

[tex]\frac{6}{0}[/tex]

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?
 
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  • #6
Why is [tex]|x| / x = -1[/tex] if x goes to + infinity? What is |x| if x becomes very large?
And my Mathematica doesn't tell me it's infinity.
 
  • #7
^^^ 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
[tex]\lim_{x \rightarrow- \infty[/tex][tex]\sqrt{9x^2+6x-5} - 3x[/tex]

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.
 

1. What does it mean to find the limit of a function as x approaches infinity?

When we say "find the limit of a function as x approaches infinity," we are essentially asking what value the function approaches as x gets larger and larger without bound. In other words, we are interested in the behavior of the function at its farthest extent to the right on the x-axis.

2. How do you find the limit of a function as x approaches infinity?

In order to find the limit of a function as x approaches infinity, we can use the rules of limits and algebraic manipulation to simplify the function. We then substitute infinity for x and evaluate the resulting expression. If the resulting value is a real number, then that is the limit of the function. If the resulting value is infinity or negative infinity, then we say that the limit does not exist.

3. What is the limit of 2x + 1 - sqrt(4x^2 + 5) as x approaches infinity?

To find the limit of this function, we can use the fact that as x gets larger and larger, the square root term becomes insignificant compared to the linear term. Therefore, we can simplify the function to 2x + 1 and as x approaches infinity, this expression approaches infinity. Therefore, the limit of 2x + 1 - sqrt(4x^2 + 5) as x approaches infinity is infinity.

4. Can a limit at infinity be a negative value?

Yes, a limit at infinity can be a negative value. This means that as x approaches infinity, the function approaches a negative value. For example, the limit of -x as x approaches infinity is negative infinity.

5. How can we use the concept of limits at infinity in real-world applications?

Limits at infinity have many real-world applications, such as in physics and economics. In physics, we can use limits at infinity to analyze the behavior of objects as they approach extreme speeds or distances. In economics, we can use limits at infinity to model the long-term behavior of a system, such as population growth or economic growth.

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