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

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Discussion Overview

The discussion revolves around finding the limit of the expression \(\frac{e^{x}-1}{\sin x}\) as \(x\) approaches 0, specifically without using L'Hospital's Rule. Participants explore various methods and mathematical concepts relevant to limits, including Taylor series and the Squeeze Theorem.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Technical explanation

Main Points Raised

  • One participant expresses uncertainty about how to approach the limit problem, mentioning the concept of punctured neighborhoods.
  • Another participant suggests rewriting the limit as \(\frac{e^x - 1}{x} \cdot \frac{x}{\sin x}\) and finding the limit of each factor separately.
  • A third participant agrees with the factorization and proposes using the Taylor series expansion for \(e^x\) to find that the limit of the first factor is 1, while considering the Squeeze Theorem for the second factor.
  • Another participant references the known limit \(\lim_{x\to 0} \frac{\sin x}{x}\) to compute the limit of the second factor by taking its reciprocal, and connects both limits to the derivatives of the functions involved at 0.

Areas of Agreement / Disagreement

Participants present multiple approaches to the problem, with no consensus on a single method being preferred. Different techniques are discussed, and while some methods are acknowledged as valid, the discussion remains open-ended without a definitive resolution.

Contextual Notes

Participants rely on various mathematical concepts such as Taylor series, the Squeeze Theorem, and derivatives, but there is no agreement on the necessity or sufficiency of these methods for solving the limit problem.

FallArk
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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,

$$\lim_{x\to 0} \frac{e^x - 1}{x} = \lim_{x\to 0} \frac{e^x - e^0}{x - 0} = \frac{d}{dx}\bigg|_{x = 0} e^x = 1$$
 

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