# Understanding a Step in Finding a Limit: Lim(x->0) x.sqrt(x+2) / sin(x)

• Cal124
In summary, understanding a step in finding a limit involves evaluating the expression Lim(x->0) x.sqrt(x+2) / sin(x). This can be done by first factoring out an x from the numerator and using the limit definition of sqrt(x) as x approaches 0. Then, using the trigonometric identity sin(x) = x as x approaches 0, the limit simplifies to Lim(x->0) sqrt(x+2) / 1, which can then be evaluated to give a final answer of 0. This step is crucial in finding the limit of the given expression and helps to simplify the calculation process.
Cal124
Apologies if this is in the wrong place. I'm struggling to understand a step in finding a limit
Lim(x->0) x.sqrt(x+2) / sin(x)
Following the given solution I get to the point where it's all divided through by x to give
Sqrt(x+2) / sin x/x
Which as approaching 0 gives
Sqrt(2) / 1 = sqrt(2)
I'm struggling how sin x/x is 1
Any help would be great and any advice on limits or links to resources would be appreciated!

The limit as sin x/x approaches 0 being 1 is usually a standard result that's done in calculus 1 courses--i.e. In most courses I've seen, the professor derives that in class, usually using the squeeze theorem, and then the students usually memorize it.

Otherwise, if you know l'Hôpital's rule, then you can use that. I'm not aware of any other methods.

Cal124
axmls said:
The limit as sin x/x approaches 0 being 1 is usually a standard result that's done in calculus 1 courses--i.e. In most courses I've seen, the professor derives that in class, usually using the squeeze theorem, and then the students usually memorize it.

Otherwise, if you know l'Hôpital's rule, then you can use that. I'm not aware of any other methods.
Thanks! Just clicked into place now, remember him explaining this now. There is some frustrating gaps in my knowledge due to a non traditional route to higher education.
Thanks again!

if you know the derivative of sin is cos, then the limit of sin(x)/x as x-->0 is the value of the derivative of sin at x=0, i.e. cos(0) = 1. of course most people prove this using the value of that limit, although there are other approaches, such as defining sin and cos as solutions of a certain differential equation, or by giving their taylor series. my detailed explanation of the limit approach is here:

http://alpha.math.uga.edu/~roy/tangents_to_y.pdf

Cal124
mathwonk said:
if you know the derivative of sin is cos, then the limit of sin(x)/x as x-->0 is the value of the derivative of sin at x=0, i.e. cos(0) = 1. of course most people prove this using the value of that limit, although there are other approaches, such as defining sin and cos as solutions of a certain differential equation, or by giving their taylor series. my detailed explanation of the limit approach is here:

http://alpha.math.uga.edu/~roy/tangents_to_y.pdf

That's a much easier way to grasp the concept, thanks! Will get reading!
Sorry, I don't suppose you have any advice/similar PDF on the definition of a limit? I feel I'm getting there (although part of me feels it could all be wrong and I'm miles away)

Cal124 said:
That's a much easier way to grasp the concept, thanks! Will get reading!
Sorry, I don't suppose you have any advice/similar PDF on the definition of a limit? I feel I'm getting there (although part of me feels it could all be wrong and I'm miles away)

You mean the episilon/delta?

MidgetDwarf said:
You mean the episilon/delta?
Yeah

## 1. What does the notation "Lim(x->0)" mean?

The notation "Lim(x->0)" represents the limit of a function as the input value approaches 0. In this case, we are finding the limit of the function x.sqrt(x+2) / sin(x) as x approaches 0.

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

To find the limit of a function, we first evaluate the function at the given input value. If the resulting value is undefined or approaches infinity, we can use algebraic manipulations or special limit rules to simplify the function and determine the limit.

## 3. Why is finding limits important?

Finding limits is important because it helps us understand the behavior of a function at a specific point. It can also help us determine if a function is continuous or if there are any discontinuities.

## 4. How do you approach finding a limit involving a square root?

When finding a limit involving a square root, we can often use the technique of rationalizing the numerator. This involves multiplying both the numerator and denominator by the conjugate of the square root term, which eliminates the radical and allows us to evaluate the limit.

## 5. Can we always use algebraic manipulations to find a limit?

No, we cannot always use algebraic manipulations to find a limit. There are some special limit rules that can be used for certain types of functions, but in some cases, the limit may not exist or may require more advanced techniques to evaluate.

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