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

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1. lim θ→0 [itex]\frac{sinθ}{θ+tanθ}[/itex]

Homework Equations



lim x→0 [itex]\frac{sinx}{x}[/itex]=1

lim x→0 [itex]\frac{cosx-1}{x}[/itex]=0

The Attempt at a Solution



lim θ→0 [itex]\frac{sinθ}{θ+sinθ/cosθ}[/itex]

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

lim θ→0 sinθ × [itex]\frac{cosθ}{θcosθ+sinθ}[/itex]

lim θ→0 [itex]\frac{θcosθ}{θcosθ+sinθ}[/itex]

The answer is supposed to be [itex]\frac{1}{2}[/itex]. What did I do wrong?
 
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physics604 said:
1. lim θ→0 [itex]\frac{sinθ}{θ+tanθ}[/itex]

Homework Equations



lim x→0 [itex]\frac{sinx}{x}[/itex]=1

lim x→0 [itex]\frac{cosx-1}{x}[/itex]=0

The Attempt at a Solution



lim θ→0 [itex]\frac{sinθ}{θ+sinθ/cosθ}[/itex]

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

lim θ→0 sinθ × [itex]\frac{cosθ}{θcosθ+sinθ}[/itex]

lim θ→0 [itex]\frac{θcosθ}{θcosθ+sinθ}[/itex]

The answer is supposed to be [itex]\frac{1}{2}[/itex]. 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.
 
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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.
 
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.
 
Nevermind, I got it! Thanks!
 
Dick said:
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!