MHB Two Similar Limits with Different Results

  • Thread starter Thread starter Velo
  • Start date Start date
  • Tags Tags
    Limits
AI Thread Summary
The discussion revolves around two similar limit exercises, one approaching positive infinity and the other negative infinity. The first limit, as x approaches positive infinity, evaluates to 1, while the second limit, as x approaches negative infinity, evaluates to -3. The confusion arises from the differing results despite the similar structure of the limits. A substitution method is suggested for the second limit to clarify the behavior as x approaches negative infinity, emphasizing the importance of handling the negative sign correctly. The conversation also touches on the implications of limits approaching specific values, such as 3, leading to undefined results.
Velo
Messages
17
Reaction score
0
So, I'm still struggling with limits a bit.. Today, I've tried solving two different exercises which look pretty much the same. I could solve the first one relatively easily:

$$\lim_{{x}\to{+\infty}}\frac{\sqrt{4x^{2}-1}-x}{x-3}$$

I applied the usual steps and arrived to the expression:

$$\lim_{{x}\to{+\infty}}\frac{3-\frac{1}{x^2}}{\sqrt{4-\frac{1}{x^2}}+1-3\sqrt{\frac{4}{x}-\frac{1}{x^3}}-\frac{3}{x}}$$

Then, by replacing x with $$+\infty$$, the answer was 1, which was correct according to my solution sheet. The second exercise is the same as above, but $$x$$ tends to $$-\infty$$ instead of $$+\infty$$. I did the same steps I used in the first exercise, and arrived at the same answer. However, the solution for the second exercise is supposed to be -3 , not 1. I'm unsure of when the $$-\infty$$ affects the expression here since it's always in the bottom part of a fraction, and so both $$+\infty$$ and $$-\infty$$ should make the fraction tend to 0. The second exercise:

$$\lim_{{x}\to{-\infty}}\frac{\sqrt{4x^{2}-1}-x}{x-3}$$

Thanks for any help in advanced!
 
Mathematics news on Phys.org
I have moved your thread here to our "Pre-Calculus" forum, as this is where we want threads regarding simple limits to be.

For the first limit, I would write:

$$\lim_{x\to+\infty}\frac{\sqrt{4x^{2}-1}-x}{x-3}=\lim_{x\to+\infty}\left(\frac{\sqrt{4x^{2}-1}-x}{x-3}\cdot\frac{\dfrac{1}{x}}{\dfrac{1}{x}}\right)=\lim_{x\to+\infty}\frac{\sqrt{4-\dfrac{1}{x^2}}-1}{1-\dfrac{3}{x}}=\frac{2-1}{1}=1$$

For the second limit, let's use the substitution:

$$u=-x$$

and the limit becomes:

$$-\lim_{u\to+\infty}\frac{\sqrt{4u^{2}-1}+u}{u+3}=-\lim_{u\to+\infty}\left(\frac{\sqrt{4u^{2}-1}+u}{u+3}\cdot\frac{\dfrac{1}{u}}{\dfrac{1}{u}}\right)=-\lim_{u\to+\infty}\frac{\sqrt{4-\dfrac{1}{u^2}}+1}{1+\dfrac{3}{u}}=-\frac{2+1}{1}=-3$$
 
Thanks for the quick reply! Sorry I created the post in the wrong forum btw, I wasn't sure where limits would go to :S

Also, hm.. Should we always use the substitution method to make the variable tend to $$+\infty$$ instead of $$-\infty$$? Or is there something about that particular limit that makes that the most viable option?
 
Another approach:

Start with

$$\lim_{x\to-\infty}\dfrac{\sqrt{4x^2-1}-x}{x-3}$$

Divide top and bottom by $x$:

$$\lim_{x\to-\infty}\dfrac{\dfrac{\sqrt{4x^2-1}}{x}-1}{1-\dfrac3x}$$

Now, to bring $x$ under the radical in the numerator, we square and take the square root but that is the absolute value of $x$, so we put a minus sign in front of the radical since $x$ tends to $-\infty$:

$$\lim_{x\to-\infty}\dfrac{-\dfrac{\sqrt{4x^2-1}}{\sqrt{x^2}}-1}{1-\dfrac3x}=-3$$
 
Another interesting limit problem specifically for this one is what if x approaches 3?
Negative infinity if $$x\to 3^-$$ and positive infinity if $$x\to 3^+$$. So if $$x\to 3$$ then DNE.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Thread 'Imaginary Pythagoras'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
Back
Top