Finding Limits: lim θ→0 \frac{sinθ}{θ+tanθ}

[physics604]

1. lim θ→0 \frac{sinθ}{θ+tanθ}

Homework Equations

lim x→0 \frac{sinx}{x}=1

lim x→0 \frac{cosx-1}{x}=0

The Attempt at a Solution

lim θ→0 \frac{sinθ}{θ+sinθ/cosθ}

lim θ→0 \frac{sinθ}{(θcosθ+sinθ)/cosθ}

lim θ→0 sinθ × \frac{cosθ}{θcosθ+sinθ}

lim θ→0 \frac{θcosθ}{θcosθ+sinθ}

The answer is supposed to be \frac{1}{2}. What did I do wrong?

 

[Dick]

1. lim θ→0 \frac{sinθ}{θ+tanθ}

Homework Equations

lim x→0 \frac{sinx}{x}=1

lim x→0 \frac{cosx-1}{x}=0

The Attempt at a Solution

lim θ→0 \frac{sinθ}{θ+sinθ/cosθ}

lim θ→0 \frac{sinθ}{(θcosθ+sinθ)/cosθ}

lim θ→0 sinθ × \frac{cosθ}{θcosθ+sinθ}

lim θ→0 \frac{θcosθ}{θcosθ+sinθ}

The answer is supposed to be \frac{1}{2}. What did I do wrong?

You haven't done anything wrong yet. You just aren't finished. Now divide numerator and denominator by θ and let θ go to zero.

 

[Student100]

Have you covered L'Hospital's rule?

 

[Dick]

Have you covered L'Hospital's rule?

Likely not, since the ingredients are the elementary trig limits.

 

[physics604]

Not in class, but I know that it's a quick way to solve limits. Meaning the derivative of the top divided by the derivative of the bottom.

 

[physics604]

I can't divide divide numerator and denominator by θ... If θ went to zero then that would make my denominator zero, which would be undefined.

 

[physics604]

Nevermind, I got it! Thanks!

 

[Student100]

Likely not, since the ingredients are the elementary trig limits.

Yeah, I should have picked up on that!

Anyways, looks like you solved their problem Dick!