Finding this limit involving sin and cos

[lo2]

Homework Statement

I have this function:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

For all $x \in R$ where $x \neq n \pi, n \in Z$

Ok I have to find the following limit:

$lim_{x\rightarrow0+}(f(x))$

Homework Equations

Limits in general and perhaps the always great Hospital's rule.

The Attempt at a Solution

I have tried to put on the same fraction line:

$f(x) = \frac{\sin{(x)}-x\cos{(x)}}{x\sin{(x)}}$

And then using the Hospital rule, but it does not really seem to bring me any further...

The first derivative of it is:

$f(x) = \frac{x^2-1+(\cos{(x)})^2}{x^2(\sin{(x)})^2}$

And then I could use the Hospital rule again but it just seems as though it will make it worse, the sinus will always be in the denominator.

 

[Curious3141]

Homework Statement

I have this function:

$f(x) = \frac{1}{x}-\frac{\cos{(x)}}{\sin{(x)}}$

For all $x \in R$ where $x \neq n \pi, n \in Z$

Ok I have to find the following limit:

$lim_{x\rightarrow0+}(f(x))$

Homework Equations

Limits in general and perhaps the always great Hospital's rule.

The Attempt at a Solution

I have tried to put on the same fraction line:

$f(x) = \frac{\sin{(x)}-x\cos{(x)}}{x\sin{(x)}}$

And then using the Hospital rule, but it does not really seem to bring me any further...

The first derivative of it is:

$f(x) = \frac{x^2-1+(\cos{(x)})^2}{x^2(\sin{(x)})^2}$

And then I could use the Hospital rule again but it just seems as though it will make it worse, the sinus will always be in the denominator.

First of all, you're not applying L'Hopital's Rule correctly. You're supposed to differentiate the numerator and denominator *separately*. Instead you differentiated the whole expression using the quotient rule. (BTW, even that expression you got can be further simplified. Use $\sin^2 x + \cos^2 x = 1$ on the numerator. Irrelevant to the question, but something you should take note of).

After you apply LHR correctly, take the reciprocal of the expression and see what it reduces to.

Alternatively, you can just apply the Taylor series throughout and get the answer quickly without using LHR.

 

[lo2]

Ah yeah ok, I can see that I have used LHR wrongly...

I think I have got the right answer applying the rule correctly! So thanks a lot :)

 

[Curious3141]

Ah yeah ok, I can see that I have used LHR wrongly...

I think I have got the right answer applying the rule correctly! So thanks a lot :)

You're welcome.