MHB What is the connection between sin and cos in Example 6.3.4?

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The discussion centers on Example 6.3.4 from "Introduction to Real Analysis," which presents the equation involving sine and cosine functions. Participants clarify that the equation does not imply the two functions are identical; rather, they share the same limit as x approaches 0 from the positive side. The application of l'Hospital's rule is highlighted as the method used to establish this limit equivalence. The conversation emphasizes understanding the nuances of limits in calculus. Overall, the example serves to illustrate the behavior of these functions near zero, not their equality.
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I am reading "Introduction to Real Analysis" (Fourth Edition) by Robert G Bartle and Donald R Sherbert ...

I am focused on Chapter 6: Differentiation ...

I need help in fully understanding the an aspect of Example 6.3.4 ...Example 6.3.4 ... reads as follows:View attachment 7305The above example implies that:

$$\frac{ \text{ sin } x}{ \sqrt{x} } = \frac{ \text{ cos } x}{ 1/ 2 \sqrt{x} } $$
Can someone please explain how/why this is true ...Peter
 
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Peter said:
The above example implies that:

$$\frac{ \text{ sin } x}{ \sqrt{x} } = \frac{ \text{ cos } x}{ 1/ (2 \sqrt{x}) } $$
Can someone please explain how/why this is true ...
No, the example does not imply that those two functions are the same. It just says that (by applying l'Hospital's rule) they have the same limit as $x\to0+$.
 
Opalg said:
No, the example does not imply that those two functions are the same. It just says that (by applying l'Hospital's rule) they have the same limit as $x\to0+$.
Oh! Of course ... how silly of me ...

Thanks Opalg ...

Peter
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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