What's the Algebraic Method to Solve the Limit of sin(x)/x?

[jkristia]

limit problem with sin(x) / ...

Homework Statement

I'm stuck trying to algebraically find the limit of this expression shown below.
If I use L'Hospital I get the answer (√3) / 2 which is the correct answer, but it seems like no matter what I try I end up with 0/0.

Any hints of how to attack this is appreciated.

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

I'm stuck trying to algebraically find the limit of this expression shown below.
If I use L'Hospital I get the answer (√3) / 2 which is the correct answer, but it seems like no matter what I try I end up with 0/0.

Any hints of how to attack this is appreciated.

Homework Equations

The Attempt at a Solution

I'm not sure why you'd want to approach it another way. Aren't you allowed to use l'Hopital? If not try substituting x=u+pi/6 and let u->0.

 

[jkristia]

Thank you for your suggestion, but even with substitution I keep going in circles. I will continue looking at this tomorow.

As for not using L'Hopital, I'm doing self study at the moment, working through problems in the book, and L'Hopital is not covered until another 7 chapters, so I figured I have to try and solve this without.

 

[Dick]

Thank you for your suggestion, but even with substitution I keep going in circles. I will continue looking at this tomorow.

As for not using L'Hopital, I'm doing self study at the moment, working through problems in the book, and L'Hopital is not covered until another 7 chapters, so I figured I have to try and solve this without.

Use the trig rule sin(a+b)=sin(a)cos(b)+cos(a)sin(b). And you know the limits of sin(u)/u and (1-cos(u))/u, yes?

 

[SammyS]

Another possibility is to write 1/2 as sin(π/6) and then use a sum to product identity from trig. It still requires you to know the limit of sin(x)/x as x → 0 .

\displaystyle \sin \theta - \sin \varphi = 2 \sin\left( \frac{\theta - \varphi}{2} \right) \cos\left( \frac{\theta + \varphi}{2} \right)

 

[jkristia]

Thank you. I think replacing 1/2 with sin(pi/6) is what I was missing. I know the limit of sin(x)/x = 1.