MHB How Do You Find the Limit of (e^x - 1) / Sin(x) as x Approaches 0?

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I'm stuck on this problem: find $$\lim_{{x}\to{0}} \frac{e^{x}-1}{\sin x}$$
Since no L'Hospital's Rule is allowed. I wonder if i can make use of the idea of punctured neighborhood(the current topic we learned in class) . Not sure how to set it up.
 
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Hi FallArk,

Write

$$\frac{e^x - 1}{\sin x} = \frac{e^x - 1}{x} \cdot \frac{x}{\sin x}$$

and find the limit of each factor on the right-hand side.
 
Euge said:
Hi FallArk,

Write

$$\frac{e^x - 1}{\sin x} = \frac{e^x - 1}{x} \cdot \frac{x}{\sin x}$$

and find the limit of each factor on the right-hand side.

Aha! that makes sense! since i can use the fact e^x = 1+x + x^2/2! +... and find that the limit is 1, for x/sinx, I am thinking squeeze rule?
 
I'm sure you've learned about the limit

$$\lim_{x\to 0} \frac{\sin x}{x}$$

in which case the limit of the second factor can be found by computing

$$\frac{1}{\lim\limits_{x\to 0} \dfrac{\sin x}{x}}$$

as $\dfrac{\sin x}{x}$ is the reciprocal of $\dfrac{x}{\sin x}$.

Really, both limits relate to derivatives of two functions at $0$. So there's no need to resort to Taylor series expansions. For example,

$$\lim_{x\to 0} \frac{e^x - 1}{x} = \lim_{x\to 0} \frac{e^x - e^0}{x - 0} = \frac{d}{dx}\bigg|_{x = 0} e^x = 1$$
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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