Two Similar Limits with Different Results

[Velo]

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:

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

I applied the usual steps and arrived to the expression:

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

Then, by replacing x with \(\displaystyle +\infty\), the answer was 1, which was correct according to my solution sheet. The second exercise is the same as above, but \(\displaystyle x\) tends to \(\displaystyle -\infty\) instead of \(\displaystyle +\infty\). 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 \(\displaystyle -\infty\) affects the expression here since it's always in the bottom part of a fraction, and so both \(\displaystyle +\infty\) and \(\displaystyle -\infty\) should make the fraction tend to 0. The second exercise:

\(\displaystyle \lim_{{x}\to{-\infty}}\frac{\sqrt{4x^{2}-1}-x}{x-3}\)

Thanks for any help in advanced!

 

[MarkFL]

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:

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

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

\(\displaystyle u=-x\)

and the limit becomes:

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

 

[Velo]

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 \(\displaystyle +\infty\) instead of \(\displaystyle -\infty\)? Or is there something about that particular limit that makes that the most viable option?

 

[Greg]

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 $-\infty$:

$$\lim_{x\to-\infty}\dfrac{-\dfrac{\sqrt{4x^2-1}}{\sqrt{x^2}}-1}{1-\dfrac3x}=-3$$

 

[suluclac]

Another interesting limit problem specifically for this one is what if x approaches 3?
Negative infinity if \(\displaystyle x\to 3^-\) and positive infinity if \(\displaystyle x\to 3^+\). So if \(\displaystyle x\to 3\) then DNE.