Calculus - taking the limit type questions

[laker_gurl3]

1.)
Take the limit as x approaches infinite...

Square root of 4x^2 + 1 / 2x-3
umm so i really don't know where to start this..i was going to multiply the numerator and denomenator by 1/x, but i don't know what the numerator would be then..

and for number 2.)

Take the limit as x approaches negative infinite:
3x/square root of x^2 + 6 (i hope you guys know wht i mean by that!)

Anyways for this one i multiplied numerator and denominator by 1/x and got a final answer of 3, but the answer in the back says -3..i just don't understand that part and where it changes to a negative..
thanks so much guys!

 

[dextercioby]

So the first is

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

HINT:Factor "2x" from both,simplify & take the limit then.

For the second,switch the limit to $+\infty$ and then do the same trick...

Daniel.

 

[Ouabache]

So the first is

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

HINT:Factor "2x" from both,simplify & take the limit then.

For the second,switch the limit to $+\infty$ and then do the same trick...

Daniel.

That is a great idea, I would also HINT that factoring out 2x from within the square root implies actually factoring out $$4x^2$$ out from the term under the root $$4x^2 +1$$

 

[dextercioby]

Yes,i thought and hoped that this itsy-bitsy detail was obvious to the OP.

Daniel.

 

[Ouabache]

For the second,switch the limit to $+\infty$ and then do the same trick...

For the 2nd part, you won't need to change the sign of the limit to $+\infty$. It is a good idea, to use the same approach as suggested for the 1st question. Here's my HINT: remember, if you factor out a term to an odd power, when you take the limit, the sign of $-\infty$ will remain, as you cancel terms from the numerator and denominator.

 

[dextercioby]

I didn't say he needed that.It was just a suggestion.I hate limits to $-\infty$.

Daniel.

 

[Ouabache]

The solution in the book is -3.
How do you find that solution with the limit $+\infty$

 

[aek]

Hey buddy, I am abiding by the physics forum rules,
SHOW ME YOUR WORK
then i might help.
:)

 

[quantumdude]

SHOW ME YOUR WORK

She did show her work. Just read the opening post.

 

[dextercioby]

The solution in the book is -3.
How do you find that solution with the limit $+\infty$

Make the substitution

$$x=-u$$

Daniel.

 

[Jameson]

I agree. Aek, you are taking your anger out on someone who was unlrelated to your ridiculous argument. Yell at the people who it involved, or better yet, if you're not going to help anyone, please just stop posting.

 

[Jameson]

Addressing the original question, couldn't you use L'Hopital's Rule? The x's have equal "power" once you apply to squareroot on top, so shouldn't the helpful rule apply when this is in indeterminant form?

 

[Ouabache]

Addressing the original question, couldn't you use L'Hopital's Rule? The x's have equal "power" once you apply to squareroot on top, so shouldn't the helpful rule apply when this is in indeterminant form?

When I try a direct application of L'Hopital's rule on Ques 1 & 2, it yields more complicated terms. So I am not picturing what you are trying to suggest.. Can you post your thought process using math?

 

[laker_gurl3]

okay thanks a lot you guys! i got the first one! still kinda working on the second one though...maybe we didn't get that far in the lesson yet but when you guys say change the negative infinite to a positive..how do i do that? and what happens to the rest of the equation?

 

[dextercioby]

$$\lim_{x\rightarrow -\infty} f(x)$$

under the change of variable

$$x=-u$$

becomes

$$\lim_{u\rightarrow +\infty} f(u)$$

Daniel.

 

[Ouabache]

As I mentioned above, for 2nd part, you won't need to change the sign of the limit to $+\infty$ You can still use the same approach you used on the 1st question.

Remember, if you factor out a term (e.g. x) which is raised to an odd power; when you take the limit, the sign of $-\infty$ will remain as you cancel terms from the numerator and denominator.

There is nothing strange about working with $-\infty$, dex just likes doing a little extra work