Final answer:
Using L'Hôpital's Rule, we can evaluate the limit by differentiating the numerator and denominator. In the given limit, the derivatives of the logarithmic numerator and the exponential denominator are found, leading to a new limit to evaluate with known properties of logarithms and exponents.
Step-by-step explanation:
When evaluating the limit of a function as x approaches infinity, specifically for the form ∞/∞ or 0/0, L'Hôpital's Rule can be employed. This rule states that if the limit of functions f(x) and g(x) as x approaches a point c is either 0/0 or ∞/∞, the limit of f(x)/g(x) as x approaches c is the same as the limit of f'(x)/g'(x) as x approaches c, where f'(x) and g'(x) are the derivatives of f(x) and g(x), respectively.
In the case of the limit Lim x→[infinity]ln(x+1)e2x−e−2x, we first note that as x approaches infinity, both the numerator and the denominator grow without bounds, possibly giving an indeterminate form of ∞/∞. To apply L'Hôpital's Rule, we take the derivative of the numerator and the derivative of the denominator. The numerator's derivative is 1/(x+1) due to the basic rules of logarithms, and the denominator's derivative follows the rules of exponents. This results in the new limit to evaluate: Lim x→[infinity](1/(x+1))/(2e2x + 2e−2x).
Using the properties of logarithms and exponents, we can simplify and evaluate the limit correctly. Remember, the logarithm of a product equals the sum of the logarithms and the logarithm of a quotient equals the difference between the logarithms.