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2.6 Reviewing Trig Functions
Before going on it might be helpful for you to remind yourself of some fundamental properties of the trigono- metric functions. There’s a good refresher here and also check out this cheat sheet. For reference throughout these notes there are some graphs.
(a) sin θ (b) cos θ (c) tan θ
Notice that − 1 ≤ sin θ ≤ 1 and − 1 ≤ cos θ ≤ 1.
2.7 Continuity of Trig Functions
We can infer the continuity of the fundamental trig functions relatively well from the above graphs.
- sin θ is continuous over all reals.
- cos θ is continuous over all reals.
For the remaining trig functions, and composites thereof, we can determine continuity using these facts and the rules from the previous section.
Example 1. Determine where tan θ is continuous.
Recall that tan θ =
sin θ cos θ
. The numerator and denominator each are continuous over all reals. Therefore this
will be discontinuous where the denominator equals zero. cos θ = 0 for rotations that take us to the y-axis.
This occurs for values θ = (2n + 1)
π 2
for any integer n.
We can repeat this process to determine the continuity of the other 3 functions.
- tan θ is continuous except when θ = (2n + 1) π 2.
- sec θ is continuous except when θ = (2n + 1) π 2.
- csc θ is continuous except when θ = (2n) π 2.
- cot θ is continuous except when θ = (2n) π 2.
Because of their nice continuity properties we can often evaluate limits of trig functions by just plugging in values.
Example 2. Find lim θ→ π 2
cos (cos θ).
By continuity we can move the limit inside.
lim θ→ π 2
cos (cos θ) = cos
cos lim θ→ π 2
θ
= cos(cos
π 2
) = cos(0) = 1
2.8 Computing Limits
Example 3. Find lim θ→∞ sin θ.
Looking at the graph, we can see that as we continue to increase the angle, sin will just oscillate back and forth from 1 to − 1 and back again. Therefore the function values are not approaching anything at all and this limit does not exist.
Example 4. Find lim y→∞ tan−^1 (y).
To evaluate this we can use the graph of tangent. Remember that inverse functions simply exchange the roll of the independent and dependent variable. To calculate this limit then we jump to the main branch of tan, pretend we’re a train on the rails, and move along towards ∞ on the y axis. What value are we approaching
on the x − axis as we do this?
π 2
Example 5. Find lim x→ 0 sin(
x
What’s going on here? The inside is going to ±∞ depending on which side we approach from. What does sin do with this? Well when x is near 0 , small changes in x give huge changes in (^) x^1. For example 1001 and 10001 differ by only. 009 bu their reciprocals differ by 900. Which in terms of radians is more than 143 rotations around the circle. So sin( (^) x^1 ) is varying wildly in values and the limit does not exist.
Figure 2.2: Graph of sin( (^1) x ) near 0.
So we’ve sandwiched the function whose limit we need to find between two others. To finish we need that the functions on the ends have limits the first we can compute and second that are equal to each other. Taking the limit as x → ∞ across the board here gives.
0 ≤ lim x→∞
sin x x
We conclude then that lim x→∞
sin x x = 0. Take a look at the graph to see this squeezing in action.
Figure 2.4: Graph of sinx^ xin red and ± (^1) x in green.
Example 8. Find lim x→ 0
x sin
x
Again we’ll use the Squeeze Theorem and start with the fact that − 1 ≤ sin (^1) x ≤ 1. If we multiply both sides by x this gives
−x ≤ x sin
x
≤ x
Taking the limit as x → 0 gives
0 ≤ lim x→ 0 x sin
x
from which we conclude that lim x→ 0 x sin
x
Figure 2.5: Graph of x sin
x
in red and ±x in green.
Example 9. Find lim θ→ 0
sin 2θ θ
The trick here will be to leverage the fact that lim x→ 0
sin x x
= 1. Remember the salient point of this identity,
since the variables are arbitrary, is that the inside of the sin function matches the denominator. So for us, since 2 θ appears inside sin we want that to also be the denominator. We can do that just fine by multiplying by in the denominator 2 as long as we also multiply the numerator by 2.
lim θ→ 0
sin 2θ θ = lim θ→ 0
sin 2θ θ = lim θ→ 0
sin 2θ 2 θ
Figure 2.8: Graph of sin(2 θ θ)in red.
Example 10. Find lim x→∞
x
)x .
There isn’t much to do here except recognize that this is the definition of e−^2.
Example 11. Find lim θ→ 0
sin θ 2
cot θ.
To start we probably want to rewrite this as lim θ→ 0
sin
θ 2
cos θ sin θ
. Now we can see that just plugging in gives
0 0 so we have work to do. I want to use the identity^ xlim→ 0
sin x x = 1 on each of the sines appearing here. So I
massage this to look how I want.
lim θ→ 0
sin θ 2 θ 2
(cos θ)
θ sin θ
