MHB Questions on Rationalizing & Limits: Answers & Explanation

[tc903]

Hi,

I tried rationalizing the first one but I get zero.

\[\lim_{x\to 3}\frac{\sqrt{19-x}-4}{\sqrt{28-x}-5}\]

\[\lim_{x\to 3}\frac{-x-3}{(\sqrt{28-x}-5)(\sqrt{19-x}+4)}\]

Here is my next question.

\[\lim_{x\to 0^{-}}\left(\frac{3}{x}-\frac{3}{|x|}\right)\]

$$f(x)=\begin{cases}-x, & x<0 \\[3pt] x, & 0\le x \\ \end{cases}$$

\[\lim_{x\to 0^{-}}\frac{6}{x}\]

I will post a picture of this later if the $\LaTeX$ didn't work.

 

[MarkFL]

Hello and welcome to MHB, tc903! :D

I tried fixing your $\LaTeX$ and I hope I interpreted things correctly.

For the first one, I suggest trying to rationalize both the numerator and the denominator simultaneously:

$$\frac{\sqrt{19-x}-4}{\sqrt{28-x}-5}\cdot\frac{\sqrt{19-x}+4}{\sqrt{19-x}+4}\cdot\frac{\sqrt{28-x}+5}{\sqrt{28-x}+5}$$

Now simplify...do you now have a determinate form?

 

[Sudharaka]

Hi,

I tried rationalizing the first one but I get zero.

\[\lim_{x\to 3}\frac{\sqrt{19-x}-4}{\sqrt{28-x}-5}\]

\[\lim_{x\to 3}\frac{-x-3}{(\sqrt{28-x}-5)\left(\sqrt{19-x}+4\right)}\]

Here is my next question.

\[\lim_{x\to 0^{-}}\left(\frac{3}{x}-\frac{3}{\left| x \right|}\right)\]

\[f(x)=\begin{cases}-x&\mbox{ when }& x<0\\
x&\mbox{ when } & x\geq 0\end{cases}\]

\[\lim_{x\to 0^{-}}\frac{6}{x}\]

I will post a picture of this later if the latex didnt work.

Hi tc903, :)

I edited your LaTeX code to make the expressions visible but I am not sure whether what I have written is what you had in mind. Let us know whether it needs additional changes.

 

[tc903]

My second question I would assume does not exist but is wrong. I did try rationalizing both the numerator and denominator beforehand. Am I correct?

\[\lim_{{x}\to 3}\frac{\sqrt{28-x}+5}{\sqrt{19-x}+4}=\frac{5}{4}\]

 

[tc903]

It wouldn't let me edit my previous post. I meant 5/4.

 

[MarkFL]

It wouldn't let me edit my previous post. I meant 5/4.

I edited your post to fix the $\LaTeX$ and change your result. That looks good to me. :D

 

[tc903]

Thanks MarkFL. I was wondering if you can clear something up with the second question though with the absolute value.

 

[MarkFL]

Thanks MarkFL. I was wondering if you can clear something up with the second question though with the absolute value.

What you did in your first post for the second question looks correct to me. You gave the piecewise definition of the absolute value and correctly simplified the limit based on that definition. Now all that is left to do is evaluate the simplified limit. :D

 

[tc903]

I stated it did not exist, but that is wrong. I substituted zero and that would be make it undefined but that is wrong. It isn't - \infty .

It is zero?

Here is a table.

x,y
-1,-6
-2,-3
-3,-2
-4,-3/2
-5,-6/5
-6,-1
-7,-6/7
-8,-3/4
-9,-2/3
-10,-3/5

 

[MarkFL]

I get $-\infty$, and W|A agrees. :D

 

[tc903]

I should have tried that. I just didnt think it was an option because it didnt say I can put it in their. The question usually asks if the limit exists find it. (type so and so for inifinity or for none exist) It had only chosen one of those. Thanks MarkFL!