To evaluate the limit

$\lim_{x \to \infty} \left(\frac{\sqrt{x^2+1}}{x+1}\right)^x,$

we'll proceed step by step:

Step 1: Simplify the base of the exponent

The expression inside the parentheses is: $\frac{\sqrt{x^2 + 1}}{x + 1}.$ Divide the numerator and denominator inside the square root by $x^2$: $\sqrt{x^2+1} = x\sqrt{1+\frac{1}{x^2}}.$ Thus, the base becomes: $\frac{\sqrt{x^2+1}}{x+1} = \frac{x\sqrt{1+\frac{1}{x^2}}}{x+1}.$ Factor $x$ out of the denominator: $x + 1 = x\left(1 + \frac{1}{x}\right).$ So: $\frac{\sqrt{x^2+1}}{x+1} = \frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}.$

Step 2: Approximation as $x \to \infty$

As $x \to \infty$, $\sqrt{1+\frac{1}{x^2}} \to 1 \quad \text{and} \quad 1+\frac{1}{x} \to 1.$ Thus: $\frac{\sqrt{x^2+1}}{x+1} \to 1.$

Step 3: Consider the exponent

The original limit is: $\lim_{x \to \infty} \left(\frac{\sqrt{x^2+1}}{x+1}\right)^x.$ Let: $y = \left(\frac{\sqrt{x^2+1}}{x+1}\right)^x.$ Take the natural logarithm of both sides: $\ln y = x \ln\left(\frac{\sqrt{x^2+1}}{x+1}\right).$ Substitute the simplified form of the base: $\ln\left(\frac{\sqrt{x^2+1}}{x+1}\right) = \ln\left(\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}\right).$

Using the approximation for $\ln(1+z) \approx z$ for small $z$, we get: $\ln\left(\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}\right) \approx \ln(1+\frac{1}{x^2}) - \ln(1+\frac{1}{x}).$ For large $x$, $\ln(1+\frac{1}{x^2}) \approx \frac{1}{x^2}$ and $\ln(1+\frac{1}{x}) \approx \frac{1}{x}$. So: $\ln\left(\frac{\sqrt{1+\frac{1}{x^2}}}{1+\frac{1}{x}}\right) \approx \frac{1}{2x^2} - \frac{1}{x}.$

Step 4: Multiply by $x$

Now: $x \ln\left(\frac{\sqrt{x^2+1}}{x+1}\right) \approx x\left(\frac{1}{2x^2} - \frac{1}{x}\right) = \frac{1}{2x} - 1.$ As $x \to \infty$, $\frac{1}{2x} \to 0$, so: $x \ln\left(\frac{\sqrt{x^2+1}}{x+1}\right) \to -1.$

Thus: $\ln y \to -1 \quad \text{and therefore} \quad y \to e^{-1}.$

Final Answer:

$\lim_{x \to \infty} \left(\frac{\sqrt{x^2+1}}{x+1}\right)^x = \frac{1}{e}.$

---

Would you like more details about this solution or specific steps? Let me know! 😊

Related Questions:

1. How does the logarithmic approximation simplify the evaluation of limits?
2. What happens to the limit if the exponent is $x^2$ instead of $x$?
3. Can L'Hôpital's Rule be applied to evaluate this limit directly?
4. How does the rate of growth of $x^2+1$ and $x+1$ affect the base of the exponent?
5. Why does $\ln(1+z) \approx z$ work well for small $z$?

Tip:

For limits involving exponents and terms that grow large, rewriting expressions using logarithms often simplifies the evaluation process.