Final answer to the problem
Step-by-step Solution
How should I solve this problem?
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- Solve using L'Hôpital's rule
- Solve without using l'Hôpital
- Solve using limit properties
- Solve using direct substitution
- Solve the limit using factorization
- Solve the limit using rationalization
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
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If we directly evaluate the limit $\lim_{x\to0}\left(\frac{\sin\left(x\right)}{\ln\left(2e^x-1\right)}\right)$ as $x$ tends to $0$, we can see that it gives us an indeterminate form
Learn how to solve limits by direct substitution problems step by step online.
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Learn how to solve limits by direct substitution problems step by step online. Find the limit of sin(x)/ln(2e^x-1) as x approaches 0. If we directly evaluate the limit \lim_{x\to0}\left(\frac{\sin\left(x\right)}{\ln\left(2e^x-1\right)}\right) as x tends to 0, we can see that it gives us an indeterminate form. We can solve this limit by applying L'Hôpital's rule, which consists of calculating the derivative of both the numerator and the denominator separately. After deriving both the numerator and denominator, and simplifying, the limit results in. Evaluate the limit \lim_{x\to0}\left(\frac{\left(2e^x-1\right)\cos\left(x\right)}{2e^x}\right) by replacing all occurrences of x by 0.
