# Calculus - taking the limit type questions

• laker_gurl3
In summary, the first problem is that the limit as x approaches infinite is not clear, and the second problem is that the answer changes when the limit is switched to +\infty .
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 teh 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 teh back says -3..i just don't understand that part and where it changes to a negative..
thanks so much guys!

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.

dextercioby said:
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 $$4x^2 +1$$

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

Daniel.

dextercioby said:
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.

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

Daniel.

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

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

aek said:

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

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

Make the substitution

$$x=-u$$

Daniel.

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.

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?

Jameson said:
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?

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 teh rest of the equation?

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

under the change of variable

$$x=-u$$

becomes

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

Daniel.

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

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## 1. What is a limit in calculus?

A limit in calculus is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It represents the value that the function approaches, but may not necessarily equal, as the input gets closer and closer to the specified value.

## 2. How do you find the limit of a function?

To find the limit of a function, you can use algebraic techniques or graphical methods. Algebraically, you can substitute the specified value for the input and see what value the function approaches. Graphically, you can observe the behavior of the function as the input approaches the specified value on a graph.

## 3. What is the difference between a one-sided and two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the specified value from one direction (either left or right). A two-sided limit takes into account both directions and requires that the function approaches the same value from both sides.

## 4. Can you take the limit of a discontinuous function?

No, you cannot take the limit of a discontinuous function at the point of discontinuity. The limit only exists if the function approaches the same value from both sides of the discontinuity, which is not possible if the function is not continuous at that point.

## 5. When is L'Hopital's rule used to evaluate limits?

L'Hopital's rule is used to evaluate limits when the limit results in an indeterminate form, such as 0/0 or ∞/∞. It allows you to simplify the function and find the limit by taking the derivative of the numerator and denominator separately.

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