Calculus I Limits

CalebCarrots

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Jan 20, 2012
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My question is on the procedure for finding the limit of trigonometric functions, specifically the problems: #1 find lim as (θ-->0) of sin(θ)/ (θ)+tan(θ). The second problem is find lim as (t-->0) of (tan6t)/(sin2t).


I am more interested in conceptualizing the information, the answers are not even necessary. Create a problem and show me the methods, and I would be greatly appreciative.:D
 
For #1, I will use x instead of theta to mitigate typing.

\(\displaystyle \displaystyle \lim_{x\to 0}\frac{sinx}{x+tanx}\)

\(\displaystyle \displaystyle =\lim_{x\to 0}\frac{sinx cosx}{xcosx+sinx}\)

Note that as \(\displaystyle x\to 0\), then \(\displaystyle sinx\approx x\) and \(\displaystyle cosx\approx 1-\frac{x^{2}}{2}\)

If we sub these into the above, we get:

\(\displaystyle \displaystyle\lim_{x\to 0}\frac{x(1-\frac{x^{2}}{2})}{x(1-\frac{x^{2}}{2})+x}=\lim_{x\to 0}\frac{x^{2}-2}{x^{2}-4}\)

Now, see the limit?.
 
For #1, I will use x instead of theta to mitigate typing.

\(\displaystyle \displaystyle \lim_{x\to 0}\frac{sinx}{x+tanx}\)

\(\displaystyle \displaystyle =\lim_{x\to 0}\frac{sinx cosx}{xcosx+sinx}\)

Note that as \(\displaystyle x\to 0\), then \(\displaystyle sinx\approx x\) and \(\displaystyle cosx\approx 1-\frac{x^{2}}{2}\)

If we sub these into the above, we get:

\(\displaystyle \displaystyle\lim_{x\to 0}\frac{x(1-\frac{x^{2}}{2})}{x(1-\frac{x^{2}}{2})+x}=\lim_{x\to 0}\frac{x^{2}-2}{x^{2}-4}\)

Now, see the limit?.

I think a student at this level would not have access to the second result. However, suppose you invert the expression. (invert it back and the end, of course)
sin x​
-------------------
x + tan x​
Inverted:
x + tan x
---------
sin x
x tan x
----- + --------
sin x sin x
sin x / cos x​
1 + -------------
sin x​
1​
1 + ------------
cos x​
1 + 1 = 2
 
Last edited:
\(\displaystyle Note: \ \ \displaystyle\lim_{x\to 0}\bigg(\frac{tanx}{x} \bigg)\ = \)

\(\displaystyle \displaystyle\lim_{x\to 0} \bigg(\frac{x}{tanx} \bigg)\ = \ 1\)

------------------------------------------



\(\displaystyle \displaystyle \lim_{x\to 0}\bigg(\frac{sinx}{x+tanx}\bigg)\)


Divide every term by tanx:


\(\displaystyle \displaystyle =\lim_{x\to 0}\bigg(\frac{cosx}{\frac{x}{tanx} + 1}\bigg)\)


\(\displaystyle = \dfrac{1}{1 + 1}\)


\(\displaystyle = \ \dfrac{1}{2}\)
 
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