Question
Determine if L'Hopital's Rule can be used. If it can, solve the limit. If it cannot choose the appropriate response. \( \lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{\sin x} \) -1 0 L'Hopital's Rule does not apply. 2
Ask by Bright Cook. in the United States
Feb 26,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
L'Hopital's Rule can be used, and the limit is 2.
Solution
We first check the limit:
limₓ→0 (eˣ – e^(–x)) = e⁰ – e⁰ = 1 – 1 = 0
limₓ→0 (sin x) = sin 0 = 0
Since we have a 0/0 indeterminate form, L'Hopital's Rule can be used.
Differentiate the numerator and the denominator:
d/dx (eˣ – e^(–x)) = eˣ + e^(–x)
d/dx (sin x) = cos x
Now, the limit becomes:
limₓ→0 (eˣ + e^(–x)) / cos x
Evaluating at x = 0:
(e⁰ + e⁰) / cos 0 = (1 + 1) / 1 = 2
Thus, the limit is 2.
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Simplify this solution The Deep Dive
To determine if L'Hopital's Rule can be used, we first check the limit along with the form of the functions involved. As \( x \) approaches 0, both the numerator \( e^{x} - e^{-x} \) and the denominator \( \sin x \) approach 0, thus yielding the indeterminate form \( \frac{0}{0} \), allowing us to apply L'Hopital's Rule. Now, applying L'Hopital's Rule involves differentiating the numerator and denominator. The derivative of the numerator, \( e^x + e^{-x} \), simplifies to \( \cosh(x) \), and the derivative of the denominator, \( \cos x \), leads to: \[ \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} = \frac{e^0 + e^0}{\cos(0)} = \frac{1 + 1}{1} = 2. \] So, the final answer is **2**.
