Math Problem Statement

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Solution

To solve the given problem, we need to determine the limit of ln(x+1+x2)\ln(x + \sqrt{1 + x^2}) as x0x \to 0.

Let's begin by examining the expression inside the logarithm.

Given: ln(x+1+x2)\ln\left(x + \sqrt{1 + x^2}\right)

As x0x \to 0, we can use the first-order Taylor expansion for small xx. For small xx, 1+x2\sqrt{1 + x^2} can be approximated as follows: 1+x21+x22\sqrt{1 + x^2} \approx 1 + \frac{x^2}{2}

So, the expression inside the logarithm becomes: x+1+x2x+1+x22x + \sqrt{1 + x^2} \approx x + 1 + \frac{x^2}{2}

Now, let's consider the logarithm: ln(x+1+x22)\ln\left(x + 1 + \frac{x^2}{2}\right)

For xx very small, we can use the Taylor expansion of ln(1+y)y\ln(1 + y) \approx y when y0y \to 0.

Here, y=x+x22y = x + \frac{x^2}{2} is also very small as x0x \to 0.

So, ln(1+x+x22)x+x22\ln\left(1 + x + \frac{x^2}{2}\right) \approx x + \frac{x^2}{2}

Thus, as x0x \to 0, ln(x+1+x2)x\ln\left(x + \sqrt{1 + x^2}\right) \sim x

Therefore, the limit is: limx0ln(x+1+x2)=x\lim_{x \to 0} \ln\left(x + \sqrt{1 + x^2}\right) = x

If you have any questions or need further details, please let me know.

Here are 5 questions you might consider asking next:

  1. Can you explain the Taylor expansion used in this solution?
  2. How would the solution change if xx approaches a different value, like infinity?
  3. Can you provide more examples of logarithmic limits?
  4. How do you approach solving limits involving square roots in general?
  5. What are some common mistakes to avoid when dealing with limits and logarithms?

Tip: When dealing with limits, using Taylor series expansions can be very helpful for approximating functions near specific points.

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Math Problem Analysis

Mathematical Concepts

Limits
Logarithms
Taylor Series

Formulas

Taylor series expansion
Logarithmic limit approximation

Theorems

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Suitable Grade Level

Advanced High School and above