# Find the limit

#### Wahamuka

##### New member
What I got in this question was positive infinity but I know its wrong. I just don’t know what the first steps are to solve this problem pls give tips

#### Attachments

• 62.5 KB Views: 7

#### Harry_the_cat

##### Senior Member
Multiply by conjugate/conjugate to get a rational expression with no sqrt on numerator.

#### Dr.Peterson

##### Elite Member
Another thing you might try is a change of variables. Consider letting x = 1/u, and finding the limit as u goes to 0 from below. This will require some care, but it's a useful trick sometimes just to try making the expression look different. (The different form will suggest the conjugate method, if you hadn't thought of it before.)

#### Wahamuka

##### New member
What did I do wrong what the heck...

#### Attachments

• 1.8 MB Views: 20

#### Dr.Peterson

##### Elite Member
When x is negative, $$\displaystyle \sqrt{x^2}\ne x$$!

#### ksdhart2

##### Senior Member
Most of your work looks good, although you made a sign error in the third line. For $$x < 0$$, we have $$\displaystyle \frac{1}{\sqrt{x^2}} = \frac{1}{|x|} = \frac{1}{-x}$$, so you (effectively) multiplied by $$\displaystyle \frac{1}{-1}$$ which necessarily changes the value of the limit.

Secondly, there's another sign error in fourth line. For $$x < 0$$, we have $$\displaystyle \sqrt{4x^2+3x}\cdot\frac{1}{x} = -\sqrt{4+\frac{3}{x}}$$.

If you fix both these sign errors, you'll find the correct answer.

#### lookagain

##### Senior Member
What I got in this question was positive infinity but I know its wrong. I just don’t know what the first steps are to solve this problem pls give tips
Also, if you know how to use the binomial theorem, what you're taking the limit of becomes

$$\displaystyle 2|x| \ + \ \dfrac{3}{4}\cdot\dfrac{x}{|x|} \ - \ \dfrac{9}{64}\cdot\dfrac{1}{|x|} \ + \ [only \ terms \ with \ the \ degree \ of \ the \ denominator \ is \ greater \ than \ 1] \ + \ 2x$$

As x approaches negative infinity, it can simplify further along to this:

the limit as x approaches negative infinity of $$\displaystyle \ \dfrac{3}{4}\cdot\dfrac{x}{|x|}$$

Continue . . .

#### Wahamuka

##### New member
I still don’t get it. What did I do wrong now?
Can you guys show your solution for this because I really don’t understand that concept of when x<0 then it the squareroot will be negative. I have another question like this I would like to solve so I just need to understand this

#### Attachments

• 1.7 MB Views: 7

#### Harry_the_cat

##### Senior Member
" I really don’t understand that concept of when x<0 then it the squareroot will be negative."
-x is not always a negative number
If say x=3, then -x = -3 but if x = -3 then -x =+3
If x is negative then -x is positive.

No-one has said that " when x<0 then it the squareroot will be negative"

In fact when talking about real numbers, the squareroot of something is always positive (or zero).

So, $$\displaystyle \sqrt {x^2} = | x |$$ and $$\displaystyle | x | = x$$ if x>0 but $$\displaystyle | x | = -x$$ if x<0.

It's like $$\displaystyle \sqrt {(-3)^2} = - -3 = 3$$.

As Dr P said in Post #5 :

#### Dr.Peterson

##### Elite Member
I still don’t get it. What did I do wrong now?
Can you guys show your solution for this because I really don’t understand that concept of when x<0 then it the squareroot will be negative. I have another question like this I would like to solve so I just need to understand this
You've corrected the errors except for one place. There should not be a "-" before the radical on the second line. Did you have a reason to put it there?

#### Wahamuka

##### New member
You've corrected the errors except for one place. There should not be a "-" before the radical on the second line. Did you have a reason to put it there?
I put the negative there because ksdhart2 said in his second to the last line post.

" I really don’t understand that concept of when x<0 then it the squareroot will be negative."
-x is not always a negative number
If say x=3, then -x = -3 but if x = -3 then -x =+3
If x is negative then -x is positive.

No-one has said that " when x<0 then it the squareroot will be negative"

In fact when talking about real numbers, the squareroot of something is always positive (or zero).

So, $$\displaystyle \sqrt {x^2} = | x |$$ and $$\displaystyle | x | = x$$ if x>0 but $$\displaystyle | x | = -x$$ if x<0.

It's like $$\displaystyle \sqrt {(-3)^2} = - -3 = 3$$.

As Dr P said in Post #5 : View attachment 12903
Also oof thats some weird typo I got

Is my solution correct now?

#### Attachments

• 2 MB Views: 10

#### Dr.Peterson

##### Elite Member
Yes, that's correct. But do you see exactly why the negative should not be there? It may help if you write out some intermediate steps.

The key part is $$\displaystyle \sqrt{4x^2+3x}\cdot\frac{1}{\sqrt{x^2}} = \sqrt{\frac{4x^2+3x}{x^2}} = \sqrt{4+\frac{3}{x}}$$. There is no place for a negative to appear from nothing!

#### Wahamuka

##### New member
Why do you apply the squareroot of x^2 = negative x only when the number you are multiplying is not in a squareroot. For example in my most recent picture, 3rd line I put (2x/-x) but I didnt put a negative before the squareroot or even inside of it

#### Dr.Peterson

##### Elite Member
I hadn't noticed that your second line there was wrong; you really multiplied by $$\displaystyle \frac{\frac{1}{\sqrt{x^2}}}{\frac{1}{\sqrt{x^2}}} = \frac{\frac{1}{-x}}{\frac{1}{\sqrt{x^2}}}$$, since for negative x, $$\displaystyle \sqrt{x^2} = |x| = -x$$. On the next line, the negative is there where it should be.

Anyway, on the third line you have replaced $$\displaystyle \sqrt{x^2}$$ with its equal, $$\displaystyle -x$$. You didn't put a negative before the remaining radical because it was never there! Why would you change the sign of anything? I showed you how to handle the radical in post #12; at no point are you replacing anything with something negative.

You seem to be thinking a little in terms of magic, as if some "rule" changes things to things they are not. No; at every step, you are replacing things only with equal quantities; each of those has its own reason for being equal, which you have to carefully consider.

#### Wahamuka

##### New member
So what even is the answer to this problem? I think im on a wild goose chase. The ans is -3/4 right?

#### lookagain

##### Senior Member
So what even is the answer to this problem? I think im on a wild goose chase. The ans is -3/4 right?

Did you look at my post # 7!?