# Limit problem not making sense

• Specter
In summary, the limit as x approaches 0 is not supposed to be a fraction, but I am having trouble getting it to work.
Specter

## Homework Statement

##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##

## The Attempt at a Solution

I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.

##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##

##\frac {lim} {x→0} \frac {(√x+1 -1)(√x+1 +1)} {x(√x+1 +1)}##

##\frac {lim} {x→0} \frac {x} {x(√x+1 +1)}##

##=\frac {1} {√1+1}##

=0.5

I have a few questions. From step 2 to step 3 the numerator gets reduced only to x. I don't know how.

When multiplying the numerator by it's conjugate, does it always result in the numerator equaling 1?

For step 4, does x just get factored out of the denominator and the numerator of x just gets changed to 1?

Specter said:

## Homework Statement

##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##

## The Attempt at a Solution

I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##

##\frac {lim} {x→0} \frac {(√x+1 -1)(√x+1 +1)} {x(√x+1 +1)}##

##\frac {lim} {x→0} \frac {x} {x(√x+1 +1)}##

##=\frac {1} {√1+1}##

=0.5

I have a few questions. From step 2 to step 3 the numerator gets reduced only to x. I don't know how.
Do you know the formula ##(a-b)\cdot (a+b) =a^2-b^2\,?## Apply it to ##a=\sqrt{x+1}## and ##b=1##.
When multiplying the numerator by it's conjugate, does it always result in the numerator equaling 1?
a.) No. b) What do you mean by ##1##? It resulted in an ##x## in this case. c) Try different values of ##a,b## in the formula above.
For step 4, does x just get factored out of the denominator and the numerator...
Yes.
... of or x just gets changed to 1?
No.

Specter
For step 4 its like having ##\frac{x}{xf(x)}## isn't that equal to ##\frac{1}{f(x)}##? The x's get simplified from the numerator and the denominator and leave just 1 behind, its like having ##\frac{x}{x}## which is 1.

So it is not x changed to 1, it is just ##\frac{x}{x}## which gets changed to its equal which is 1.

Specter
Specter said:

## Homework Statement

##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.
As regards the LaTeX ...

To get the limit to display correctly use:
##\lim_{x \to 0} \frac {\sqrt {x+1}-1} {x} ## which gives ##\lim_{x \to 0} \frac {\sqrt {x+1}-1} {x} ##

OR use \displaystyle to improve readability.

##\displaystyle \lim_{x \to 0} \frac {\sqrt {x+1}-1} {x} ## which gives ##\displaystyle \lim_{x \to 0} \frac {\sqrt {x+1}-1} {x} ##

Specter and YoungPhysicist
If you aren't happy with the result, check by doing it another way. Try expanding (1+x)1/2 as a Taylor series.

@Specter :
Rather than simply marking the thread as being solved, please respond to the replies to the thread.

It's nice that you "Like" the responses of @Delta2 and @fresh_42 , but replies showing that you actually understand their posts would be a much better demonstration of showing that you appreciate the effort they put forth to respond to your questions.

Last edited:
epenguin
Specter said:

## Homework Statement

##\frac {lim} {x→0} \frac {\sqrt {x+1}-1} {x}##
I know the limit as x approaches 0 isn't supposed to be a fraction but I can't get the x approaches 0 under the lim.

I couldn't get some of this typed out in latex, it just wouldn't work for me. But here's the rest of the work. The square root is only over the x+1's.

Another method you may want to try is L'Hopital's Rule which is an efficient method of calculating some of these limits questions when you have fractions...

L'Hopitals Rule states that
$$\displaystyle \lim_{x \to c} \frac {f(x)} {g(x)} = \displaystyle \lim_{x \to c} \frac {f^n(x)} {g^n(x)}$$

where n is the nth derivative...

so you could differentiate the numerator and denominator independently and form the new fraction. Then you can plug zero into get the answer. Sometimes you may have to differentiate again as well.

Hope that is of some help. Just wanted to provide another way of tackling these problems.

Delta2
fresh_42 said:
Do you know the formula ##(a-b)\cdot (a+b) =a^2-b^2\,?## Apply it to ##a=\sqrt{x+1}## and ##b=1##.

a.) No. b) What do you mean by ##1##? It resulted in an ##x## in this case. c) Try different values of ##a,b## in the formula above.

Yes.No.
Thank you, this helped me make some sense of it.

Delta2

## 1. What is a limit problem?

A limit problem is a mathematical concept that involves finding the value that a function approaches as its input variable gets closer and closer to a specific value. It is used to describe the behavior of a function near a particular point.

## 2. Why do limit problems sometimes not make sense?

Limit problems may not make sense if the function is undefined or has a discontinuity at the point being evaluated. In these cases, the limit does not exist and the problem cannot be solved.

## 3. How can I determine if a limit problem makes sense?

If the function is continuous at the point being evaluated, then the limit problem will make sense. This means that the function has no breaks or gaps and can be drawn without lifting the pencil from the paper.

## 4. Can I use algebra to solve limit problems?

Yes, algebraic manipulation can be used to solve many limit problems. However, there are some cases where other methods, such as L'Hopital's rule or graphing, may be necessary.

## 5. Are there any common mistakes to avoid when solving limit problems?

One common mistake is to assume that the limit exists without checking if the function is continuous at the point being evaluated. It is also important to check if the limit is a one-sided or two-sided limit and to use the correct notation when writing the answer.

• Calculus and Beyond Homework Help
Replies
1
Views
557
• Calculus and Beyond Homework Help
Replies
13
Views
2K
• Calculus and Beyond Homework Help
Replies
8
Views
891
• Calculus and Beyond Homework Help
Replies
6
Views
444
• Calculus and Beyond Homework Help
Replies
4
Views
625
• Calculus and Beyond Homework Help
Replies
5
Views
422
• Calculus and Beyond Homework Help
Replies
10
Views
956
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
956
• Calculus and Beyond Homework Help
Replies
6
Views
968