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Find the limits.$$\lim _{x \rightarrow 0} \frac{e^{x}-1}{\sin x}$$
$$\lim _{x \rightarrow 0} \frac{e^{x}-1}{\sin x}=1$$
Calculus 1 / AB
Chapter 3
TOPICS IN DIFFERENTIATION
Section 6
L Hopital's Rule; Indeterminate Forms
Functions
Limits
Derivatives
Differentiation
Continuous Functions
Applications of the Derivative
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So we're looking at the limits. This x goes to zero of e to the X minus one divided by sine of axe. Now let's first try Teoh. Uh, just do this limit on the numerator. We get zero on the denominator. We get zero with 00 That's an indeterminant form. So we are good to go to use low petals rule. So let's apply low bottles rule. And that's next step. I like to write a little l h over the equal sign just to indicate that we're doing low petals rule. So we take the derivative of the numerator that's e to the X. Take the derivative of the denominator That's co sign. And, uh, now we can do this limit. We get 1/1 has won our answer
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