Limit x->inf of: (ln(sqrt(x+1)+4)/(ln(sqrt(7x-2)+3)

To find the limit of the given expression as \( x \) approaches infinity, we will simplify and analyze the expression step by step.

The expression is:

\[
\lim_{x \to \infty} \frac{\ln\left(\sqrt{x+1} + 4\right)}{\ln\left(\sqrt{7x-2} + 3\right)}
\]

### Step 1: Approximate the logarithms
For large \( x \), the terms \( \sqrt{x+1} + 4 \) and \( \sqrt{7x-2} + 3 \) can be approximated by their leading terms.

- \( \sqrt{x+1} \approx \sqrt{x} \), so \( \sqrt{x+1} + 4 \approx \sqrt{x} + 4 \).
- \( \sqrt{7x-2} \approx \sqrt{7x} = \sqrt{7}\sqrt{x} \), so \( \sqrt{7x-2} + 3 \approx \sqrt{7}\sqrt{x} + 3 \).

### Step 2: Simplify the logarithms

Using the approximation in large \( x \):

\[
\ln\left(\sqrt{x} + 4\right) \approx \ln\left(\sqrt{x}(1 + \frac{4}{\sqrt{x}})\right) = \ln\sqrt{x} + \ln\left(1 + \frac{4}{\sqrt{x}}\right)
\]

\[
\ln\sqrt{x} = \frac{1}{2} \ln x
\]

For large \( x \), \( \frac{4}{\sqrt{x}} \) is small, so we can use the approximation \( \ln(1 + u) \approx u \) when \( u \) is small:

\[
\ln\left(\sqrt{x} + 4\right) \approx \frac{1}{2} \ln x + \frac{4}{\sqrt{x}}
\]

Similarly:

\[
\ln\left(\sqrt{7x} + 3\right) \approx \frac{1}{2} \ln(7x) + \frac{3}{\sqrt{7x}}
\]

\[
\ln\left(\sqrt{7x} + 3\right) = \frac{1}{2} (\ln 7 + \ln x) + \frac{3}{\sqrt{7x}} = \frac{1}{2} \ln x + \frac{1}{2} \ln 7 + \frac{3}{\sqrt{7x}}
\]

### Step 3: Take the limit

Now, substituting these approximations back into the limit:

\[
\lim_{x \to \infty} \frac{\ln\left(\sqrt{x} + 4\right)}{\ln\left(\sqrt{7x} + 3\right)} = \lim_{x \to \infty} \frac{\frac{1}{2} \ln x + \frac{4}{\sqrt{x}}}{\frac{1}{2} \ln x + \frac{1}{2} \ln 7 + \frac{3}{\sqrt{7x}}}
\]

For large \( x \), the terms \( \frac{4}{\sqrt{x}} \) and \( \frac{3}{\sqrt{7x}} \) become negligible, so the limit simplifies to:

\[
\lim_{x \to \infty} \frac{\frac{1}{2} \ln x}{\frac{1}{2} \ln x + \frac{1}{2} \ln 7} = \lim_{x \to \infty} \frac{\ln x}{\ln x + \ln 7}
\]

### Step 4: Final Simplification

We can divide the numerator and denominator by \( \ln x \):

\[
\lim_{x \to \infty} \frac{1}{1 + \frac{\ln 7}{\ln x}} 
\]

As \( x \) approaches infinity, \( \frac{\ln 7}{\ln x} \) approaches 0, so the limit simplifies to:

\[
\lim_{x \to \infty} \frac{1}{1 + 0} = 1
\]

### Final Answer

\[
\boxed{1}
\]

Do you want any more details or have any questions?

---

Here are some related questions:
1. How would the limit change if the constants inside the logarithms were different?
2. What if \( x \) approaches zero instead of infinity? How would the analysis differ?
3. How does L'Hôpital's rule apply to logarithmic limits of this form?
4. Can this method be extended to logarithms with higher-degree polynomials?
5. How would this limit behave if we replaced \( \ln(x) \) with \( \log_b(x) \) for any base \( b \)?

**Tip:** Always consider simplifying expressions with approximations when dealing with limits at infinity.

Limit x->inf of: (ln(sqrt(x+1)+4)/(ln(sqrt(7x-2)+3))

Explore the detailed solution for finding the limit as x approaches infinity for the expression ln(sqrt(x+1) + 4) / ln(sqrt(7x-2) + 3). This problem covers logarithmic approximations and asymptotic behavior, with step-by-step guidance for simplifying and solving. Suitable for advanced high school or early college students.

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