# Finding the limit of trignometric functions

In summary, the conversation discusses finding the limit of a given function using relevant equations and breaking up the limit of a product. The final answer is determined to be 2, with consideration for L'Hopital's Rule for a more efficient solution.

## Homework Statement

Find the Limit: lim h->0 of ( h sin h)/ ( 1-cos h)

## Homework Equations

lim h->0 of (sin theta)/ theta = 1
lim h->0 of (cos(h)-1)/h=0

## The Attempt at a Solution

lim h->0 (h/1-cos h)*(sin h) * (1+cos h/ 1 +cos h)

= lim h->0 (h(1+cos h) * sin h)/(1-cos^2 h)

= lim h->0 (h(1+cos h))/sin h

That's how far I got, I have no idea how to continue.

## Homework Statement

Find the Limit: lim h->0 of ( h sin h)/ ( 1-cos h)

## Homework Equations

lim h->0 of (sin theta)/ theta = 1
lim h->0 of (cos(h)-1)/h=0

## The Attempt at a Solution

lim h->0 (h/1-cos h)*(sin h) * (1+cos h/ 1 +cos h)

= lim h->0 (h(1+cos h) * sin h)/(1-cos^2 h)

= lim h->0 (h(1+cos h))/sin h

That's how far I got, I have no idea how to continue.
You're almost there!
$$=\lim_{h \to 0} \frac{h}{sin(h)}\cdot (1 + cos(h))$$

You can break up the limit of a product to the product of limits, provided that both limits exist. Hint: look at your relevant equations.

Mark44 said:
You're almost there!
$$=\lim_{h \to 0} \frac{h}{sin(h)}\cdot (1 + cos(h))$$

You can break up the limit of a product to the product of limits, provided that both limits exist. Hint: look at your relevant equations.

Okay, so I split them up, but would the equation: lim h->0 of sin h/ h = 1 be the same as the lim h->0 of h/sin h?? how does that work?

and therefore be: 1 * lim h->0 (1+ cos h)
= 1 * 2 = 2?

Yup, I get 2 also.

If you know limh→0 h/sinh = 1 then you know that
limh→0 (sinh/h)-1 = 1.

But (sinh/h)-1 = 1 ⇒ (sinh/h)= 1

Edit: You don’t know l’hopital’s rule right? If you do this is a rather round-about way to do this problem.

Last edited:
JonF said:
Yup, I get 2 also.
Agreed.
JonF said:
If you know limh→0 h/sinh = 1 then you know that
limh→0 (sinh/h)-1 = 1.

But (sinh/h)-1 = 1 ⇒ (sinh/h)= 1
Not sure what the missing symbol is - implies?
But sinh/h $\neq$ 1. The limit as h -> 0 is 1, though.
JonF said:
Edit: You don’t know l’hopital’s rule right? If you do this is a rather round-about way to do this problem.
But if the OP doesn't know L'Hopital's Rule, this is the way to go.

Yah, I haven't learned L'Hopital's Rule yet

But thanks guys!

## What is the definition of a limit?

The limit of a function is the value that the function approaches as its input (x) approaches a certain value (a).

## How do you find the limit of a trigonometric function?

To find the limit of a trigonometric function, you can use algebraic manipulations, trigonometric identities, and the properties of limits such as the limit laws and squeeze theorem.

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

A one-sided limit only considers the behavior of the function as the input approaches the given value from one direction (left or right), while a two-sided limit considers the behavior of the function from both directions.

## Can the limit of a trigonometric function be undefined?

Yes, the limit of a trigonometric function may be undefined if the function oscillates or approaches different values from different directions as the input approaches a certain value.

## What are some real-world applications of finding the limit of trigonometric functions?

Finding the limit of trigonometric functions is important in fields such as physics, engineering, and economics. It can be used to model and predict the behavior of systems and functions in the real world, such as pendulum motion, sound waves, and economic trends.

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