Calculus I Limits

[CalebCarrots]

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.

 

[galactus]

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?.

 

[SAMUELK]

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

 

[lookagain]

$\displaystyle Note: \ \ \displaystyle\lim_{x\to 0}\bigg(\frac{tanx}{x} \bigg)\ =$

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

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



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


Divide every term by tanx:


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


$\displaystyle = \dfrac{1}{1 + 1}$


$\displaystyle = \ \dfrac{1}{2}$