Solve [ASK] Limit of Cosine: How to?

  • Context: MHB 
  • Thread starter Thread starter Monoxdifly
  • Start date Start date
  • Tags Tags
    Cosine Limit
Click For Summary

Discussion Overview

The discussion revolves around evaluating the limit $$\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos{x}}{x-\frac{\pi}{2}}$$. Participants explore various approaches to solve this limit, including substitutions and the application of trigonometric identities. The scope includes mathematical reasoning and limit evaluation techniques.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant initially attempts to convert $$\cos x$$ to $$\frac{\tan{x}}{\sin{x}}$$ but questions the applicability of this approach.
  • Another participant suggests a substitution $$u=\frac{\pi}{2}-x$$ to reformulate the limit.
  • Following the substitution, a participant expresses the limit as $$\lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$$.
  • There is a discussion about factoring the negative sign in the denominator and applying a co-function identity to the numerator.
  • Concerns are raised about encountering the indeterminate form $$\frac{0}{0}$$ in the limit expression involving $$\cos{\frac{\pi}{2}}$$.
  • Another participant recalls the co-function identity $$\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$$ and reformulates the limit as $$L=-\lim_{u\to0}\left(\frac{\sin(u)}{u}\right)$$.
  • There is acknowledgment of the indeterminate form and reference to the well-known limit $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$$.
  • A participant introduces a geometric argument using the unit circle to establish bounds for $$\frac{\sin(x)}{x}$$ and applies the squeeze theorem to conclude that $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$$.

Areas of Agreement / Disagreement

Participants express various approaches and reasoning, but there is no consensus on a final solution or method for evaluating the limit. The discussion remains exploratory with multiple perspectives presented.

Contextual Notes

Some participants note the presence of indeterminate forms and the reliance on specific limit results, which may depend on prior knowledge or assumptions about trigonometric limits.

Monoxdifly
MHB
Messages
288
Reaction score
0
How to solve $$\lim_{x\rightarrow\frac{\pi}{2}}\frac{\cos{x}}{x-\frac{\pi}{2}}$$? At first I tried to convert cos x to $$\frac{\tan{x}}{\sin{x}}$$ but then realized that $$\lim_{x\rightarrow c}\frac{\tan{x}}{x}$$ only applies if c = 0. So, how?
 
Physics news on Phys.org
Let's use the substitution:

$$u=\frac{\pi}{2}-x$$

What does the limit become then?
 
$$\lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$$?
 
Monoxdifly said:
$$\lim_{u\rightarrow0}\frac{\cos{(\frac{\pi}{2}-u)}}{-u}$$?

Yes, and so I would factor the negative sign in the denominator out in front of the limit, and apply a co-function identity to the numerator...(Thinking)
 
Won't it give $$\frac{0}{0}$$ again in the $$\frac{\cos{\frac{\pi}{2}}\cdot\cos{u}}{u}$$ part?
 
Monoxdifly said:
Won't it give $$\frac{0}{0}$$ again in the $$\frac{\cos{\frac{\pi}{2}}\cdot\cos{u}}{u}$$ part?

Recall the co-function identity:

$$\cos\left(\frac{\pi}{2}-\theta\right)=\sin(\theta)$$

And so our limit becomes:

$$L=-\lim_{u\to0}\left(\frac{\sin(u)}{u}\right)$$

And yes, this is an indeterminate form, but it is well-known to students and usually allowed as a result to use:

$$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$$
 
Oops... Sorry sorry... Totally forgot about that...
 
Consider the following diagram of the unit circle in the first quadrant:

View attachment 6091

From this, we can see that for $0\le x\le\dfrac{\pi}{2}$ we have:

$$\sin(x)\le x\le\tan(x)$$

Dividing though by $\sin(x)$, we obtain:

$$1\le\frac{x}{\sin(x)}\le\frac{1}{\cos(x)}$$

or:

$$\cos(x)\le\frac{\sin(x)}{x}\le1$$

And so we must then have:

$$\lim_{x\to0}\left(\cos(x)\right)\le\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\le\lim_{x\to0}(1)$$

or:

$$1\le\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)\le1$$

Hence (by the squeeze theorem):

$$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)=1$$
 

Attachments

  • squeeze.png
    squeeze.png
    1.5 KB · Views: 129

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 29 ·
Replies
29
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K