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How do you find the limit of ln(2x)−ln(1+x)
as x approaches infinity?

User Reaz
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1 Answer

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Final answer:

The limit of ln(2x) - ln(1+x) as x approaches infinity is the natural logarithm of 2, which is approximately 0.693.

Step-by-step explanation:

To find the limit of ln(2x) - ln(1+x) as x approaches infinity, we can use logarithmic properties and the concept of limits. First, we use the logarithm property that allows us to combine the two logarithms into a single logarithmic expression, as follows:

ln(2x) - ln(1+x) = ln(\frac{2x}{1+x}).

As x approaches infinity, the expression \frac{2x}{1+x} approaches 2 since the presence of 1 becomes negligible compared to the infinitely large x. Thus, the expression inside the logarithm approaches 2. Now, we apply the limit:

lim_(x \to \infty) ln(\frac{2x}{1+x}) = ln(lim_(x \to \infty) \frac{2x}{1+x}) = ln(2).

To find the limit of ln(2x) - ln(1+x) as x approaches infinity, we can simplify the expression using logarithmic properties. First, we can combine the logarithms into a single logarithm by using the property ln(a) - ln(b) = ln(a/b). So, ln(2x) - ln(1+x) = ln((2x)/(1+x)).

Next, we can divide the numerator and denominator by x, resulting in ln(2/(1/x + 1)). As x approaches infinity, 1/x approaches 0, so the denominator approaches 1. Therefore, the expression simplifies to ln(2/1) = ln(2).

Thus, the limit of ln(2x) - ln(1+x) as x approaches infinity is ln(2).

The result is that the limit of the original expression as x approaches infinity is ln(2), which is a constant approximately equal to 0.693.

User Kisaan
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