Math Problem Statement

Find the limit ((√(x)+2)/(x+3))^(x) when x tends to infinity

Solution

To solve limx(x+2x+3)x\lim_{x \to \infty} \left( \frac{\sqrt{x} + 2}{x + 3} \right)^x, let's proceed step-by-step.


Step 1: Express the function as elne^{\ln}

The limit can be rewritten using the exponential and natural logarithm: limx(x+2x+3)x=exp(limxxln(x+2x+3)).\lim_{x \to \infty} \left( \frac{\sqrt{x} + 2}{x + 3} \right)^x = \exp\left(\lim_{x \to \infty} x \ln\left(\frac{\sqrt{x} + 2}{x + 3}\right)\right).


Step 2: Simplify the logarithmic term

Let: y=ln(x+2x+3)=ln(x+2)ln(x+3).y = \ln\left(\frac{\sqrt{x} + 2}{x + 3}\right) = \ln(\sqrt{x} + 2) - \ln(x + 3).

For large xx, approximate each term:

  1. ln(x+2)=ln(x(1+2x))=lnx+ln(1+2x)\ln(\sqrt{x} + 2) = \ln(\sqrt{x}(1 + \frac{2}{\sqrt{x}})) = \ln\sqrt{x} + \ln(1 + \frac{2}{\sqrt{x}}).
    Using ln(1+u)u\ln(1 + u) \approx u for small uu, ln(1+2x)2x\ln(1 + \frac{2}{\sqrt{x}}) \approx \frac{2}{\sqrt{x}}.
    Thus, ln(x+2)12lnx+2x\ln(\sqrt{x} + 2) \approx \frac{1}{2} \ln x + \frac{2}{\sqrt{x}}.

  2. ln(x+3)=ln(x(1+3x))=lnx+ln(1+3x)\ln(x + 3) = \ln(x(1 + \frac{3}{x})) = \ln x + \ln(1 + \frac{3}{x}).
    Using ln(1+u)u\ln(1 + u) \approx u for small uu, ln(1+3x)3x\ln(1 + \frac{3}{x}) \approx \frac{3}{x}.
    Thus, ln(x+3)lnx+3x\ln(x + 3) \approx \ln x + \frac{3}{x}.

Now, the difference becomes: ln(x+2x+3)(12lnx+2x)(lnx+3x)=12lnx+2x3x.\ln\left(\frac{\sqrt{x} + 2}{x + 3}\right) \approx \left(\frac{1}{2} \ln x + \frac{2}{\sqrt{x}}\right) - \left(\ln x + \frac{3}{x}\right) = -\frac{1}{2} \ln x + \frac{2}{\sqrt{x}} - \frac{3}{x}.


Step 3: Multiply by xx and analyze the dominant term

xln(x+2x+3)x(12lnx+2x3x).x \ln\left(\frac{\sqrt{x} + 2}{x + 3}\right) \approx x\left(-\frac{1}{2} \ln x + \frac{2}{\sqrt{x}} - \frac{3}{x}\right).

  1. The dominant term here is 12xlnx-\frac{1}{2} x \ln x, as it grows faster than the other terms (2xx=2x\frac{2x}{\sqrt{x}} = 2\sqrt{x} and 3-3).
  2. Thus, xln(x+2x+3)x \ln\left(\frac{\sqrt{x} + 2}{x + 3}\right) \to -\infty as xx \to \infty.

Step 4: Conclude the limit

exp(limxxln(x+2x+3))=exp()=0.\exp\left(\lim_{x \to \infty} x \ln\left(\frac{\sqrt{x} + 2}{x + 3}\right)\right) = \exp(-\infty) = 0.

The limit is: 0.\boxed{0}.


Would you like further clarifications or a breakdown of specific steps?


Related Questions:

  1. What is the growth rate comparison of lnx\ln x and x\sqrt{x}?
  2. How does the expansion of logarithmic terms affect limit problems?
  3. Can L'Hôpital's rule be applied to this problem? Why or why not?
  4. What are the common dominant term approximations for limits at infinity?
  5. How does the natural logarithm transform exponential limits?

Tip: When analyzing limits involving exponentials, focus on the dominant terms in logarithmic expansions.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Limits
Logarithmic Transformations
Exponential Functions
Asymptotic Approximations

Formulas

e^ln(x) = x
ln(a/b) = ln(a) - ln(b)
ln(1 + u) ≈ u (for small u)

Theorems

Properties of Exponents and Logarithms
Dominant Term Analysis for Limits at Infinity

Suitable Grade Level

Grades 11-12 and early college