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

[FallArk]

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.

 

[Euge]

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.

 

[FallArk]

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?

 

[Euge]

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,

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