How can I evaluate limits involving trig functions?

In summary, lim as x->0 of (sin 3x)/2x, lim as x->0 of (tan 5x)/(sin 2x), lim as x->0 of (sin²3x)/2x, and lim as h->0 of [(h+x)³ -x³]/h all have limits as x goes to 0 that are sin(x)/sin(x). However, for (sin²3x)/2x, there is no sin² button on the calculator, so the user has to use sin3x*sin3x/2x to find the limit.
  • #1
burge
5
0
I am having trouble with the following problems:

1) lim as x -> 0 of (sin 3x)/2x

2) lim as x -> 0 of (tan 5x)/(sin 2x)

3) lim as x -> 0 of (sin²3x)/2x

4) lim as h -> 0 of [(h+x)³ -x³]/h

5) lim as h -> 0 of [1/(x+h) - 1/x]/h


*Thanks for your help
 
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  • #2
These are basic formulas

for eg u can write sinx~x when x->0 similarly for tanx

What Have u covered so far in Limits this will help us to give u better explanations
 
  • #3
We know that sin and cos are continuous,
the lim as x-> 0 (sin (x))/x = 1
the lim as x-> 0 (1 - cos(x))/x = 0
and some basic trig identities
 
  • #4
It may be that your problem is tjat "(sin 3x)/2x" has different coefficients for the x inside and outside of the sine. That no big deal. First take out the "1/2": (1/2)(sin3x)/x and the multiply both numerator and denominator by 3: (3/2)(sin3x)/3x Think of the "3x" as u and the problem is (3/2) (sin u)/u. Since u= 3x goes to 0 when x does, what is the limit of that as u goes to 0?
 
  • #5
I got that one now, I'm really stuck on (sin² 3x)/2x
Also, I know there is a trick for finding it in the calculator, since there is no sin² button, but I don't remember it.
 
  • #6
write it as sin3x*sin3x/2x and now find the limit it would be zero
 
  • #7
1) lim as x -> 0 of (sin 3x)/2x



set u = 3x as halls of ivy suggests



2) lim as x -> 0 of (tan 5x)/(sin 2x)

tan(5x) = sin(5x)/cos(5x) and it suffices to find out what sin(5x)/sin(2x) tends to so you could work it all out in terms of sin2x

or look at the end of the post



3) lim as x -> 0 of (sin²3x)/2x


4) lim as h -> 0 of [(h+x)³ -x³]/h

expand the bracket


5) lim as h -> 0 of [1/(x+h) - 1/x]/h

have you done basic algebra? cos that again is 'just do the manipulation' - think before asking, you'll learn a lot more


in general you want to use l'Hopital's rule.
 

1. What is a limit with a trig function?

A limit with a trig function refers to the mathematical concept of determining the value that a trigonometric function approaches as the input variable approaches a certain value. It is used to find the behavior of a trigonometric function near a specific point.

2. How do you evaluate a limit with a trig function?

To evaluate a limit with a trig function, you can use algebraic manipulation, trigonometric identities, and the properties of limits. You can also use a graphing calculator or a table of values to estimate the limit.

3. What are the common trigonometric limits?

The common trigonometric limits include:

  • sin(x)/x as x approaches 0
  • cos(x)-1/x as x approaches 0
  • tan(x)/x as x approaches 0
  • sin(x)/x as x approaches infinity
  • 1-cos(x)/x as x approaches 0
  • 1-cos(x)/x as x approaches infinity

4. Can limits with trig functions have different values from the left and right?

Yes, limits with trig functions can have different values from the left and right. This is known as a one-sided limit. If the limit from the left and right is equal, then the limit exists. However, if the limit from the left and right is not equal, then the limit does not exist.

5. How do you prove the existence of a limit with a trig function?

To prove the existence of a limit with a trig function, you can use the epsilon-delta definition of a limit. This involves showing that for any small number ε, there exists a corresponding number δ, such that if the input variable is within δ units of the limit point, then the output of the function will be within ε units of the limit value.

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