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.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top