I have a midterm test 2moro and i

In summary, to find the limit of x cos x-sin x/x sin^2 x as x approaches 0, we can use L'Hopital's rule. First, we must show that both the numerator and denominator go to 0 as x goes to 0. Then, we can take the derivatives of the numerator and denominator and apply L'Hopital's rule again until we have a limit that can be solved without it. In this case, the final limit is -1/3.
  • #1
HELP!!!
2
0

Homework Statement


lim x cos x-sin x/x sin^2 x
x-0



Homework Equations





The Attempt at a Solution


............
no clue
i just know that we should use l'hopitals rule
 
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  • #2
L'Hopital's rule states that if you have a limit of a fraction where both the numerator and denominator of the fraction go to 0 or infinity (in this case infinity), the limit is equal to the same limit of the derivatives of the numerator and denominator.

So in your particular limit, let's assume that f(x)=x*cosx-sinx and g(x)=s*sin^2 x

Then first you must show that both the numerator and denominator go to 0 as x goes to 0.

lim x*cosx-sinx
x->0

This goes to 0 because as x goes to 0, x*cosx goes to 0, and because sin(0) is 0, so does sinx.

lim x*sin^2 x
x->0

This also goes to 0 because sin goes to 0 as well as x.

So because of this, L'Hopital's rule then states:

lim x*cosx-sinx/x*sin^2 x = lim f(x)/g(x) = lim f'(x)/g'(x) (all of these limits are as x->0)


so we then take the derivatives of f(x) and g(x).

Using the product rule we find that:
f'(x) = cosx-x*sinx-cosx = -x*sinx

Using the same rule we find that:
g'(x) = 2*x*cosx*sinx+sin^2 x

So the limit is now:
lim -x*sinx/2*x*cosx*sinx+sin^2 x
x->0

canceling sinx makes:
lim -x/2*x*cosx+sinx
x->0

Now if you look at this new limit, you can see that the top and bottom once again both go to 0. So L'Hopital's rule can be applied once again:

f(x)=-x
g(x)=2*x*cosx+sinx

it is easily seen that f'(x)=-1

From the product and addition rule we see that:
g'(x)=2*cosx-2*x*sinx+cosx = 3*cosx-2*x*sinx

So our new (and final) limit is:

lim -1/3*cosx-2*x*sinx
x->0

Now, in this limit, we only need to break up the terms.

2*x*sinx will go to 0, so we can cancel that out, however, 1, and 3*cosx do not.

cos(0)=1, so 3*cosx as x->0 is equal to 3.

So in the end we have the following limit:

lim -1/3*cosx
x->0

And because 1 goes to 1, and 3*cosx goes to 3, we obtain our final answer:

lim x*cosx-sinx/x*sin^2 x = -1/3
x->0

Sorry that I didn't use LaTex for the equations. So if they look a little unclear in their format, writing them down will help.
 
Last edited:

Related to I have a midterm test 2moro and i

1. What should I study for the midterm test tomorrow?

You should study all the material that has been covered in class so far. Be sure to review your notes, textbook, and any study guides provided by your professor.

2. How should I prepare for the midterm test?

Make a study schedule and stick to it. Practice with old tests or quizzes, and make sure to get a good night's sleep before the test. Also, try to reduce stress by staying organized and taking breaks when needed.

3. What if I didn't do well on the first midterm?

Don't panic. Use this as a learning opportunity and focus on areas where you struggled. Talk to your professor or a tutor for extra help, and make sure to put in extra effort for the second midterm.

4. Should I study alone or with a group?

This depends on your learning style. If you prefer studying alone, then that may be the best option for you. However, if you benefit from discussing and explaining concepts to others, then a study group may be helpful.

5. What should I do if I feel overwhelmed or anxious about the test?

Take a deep breath and try to relax. Remember that it is normal to feel nervous before a test. Make sure to take care of yourself by getting enough rest, eating well, and taking breaks when needed. If you continue to feel overwhelmed, reach out to a friend, family member, or a counselor for support.

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