For `x in R, underset(x to oo)lim ((x-3)/(x+2))^(x)=`

Options

A. e
B. `e^(-1)`
C. `e^(-5)`
D. `e^(5)`

Answer

To solve the limit \(\lim_{x \to \infty} \left(\frac{x-3}{x+2}\right)^{x}\), we can follow these steps:

### Step 1: Identify the form of the limit
As \(x\) approaches infinity, both the numerator and denominator approach infinity, which gives us the indeterminate form \(\frac{\infty}{\infty}\). Hence, we can rewrite the expression in a more manageable form.

### Step 2: Rewrite the limit
We can express the limit as:
\[ \lim_{x \to \infty} \left(\frac{x-3}{x+2}\right)^{x} = \lim_{x \to \infty} e^{x \ln\left(\frac{x-3}{x+2}\right)} \]
This transformation is useful because it allows us to analyze the exponent separately.

### Step 3: Simplify the logarithm
Now we need to simplify the logarithm:
\[ \ln\left(\frac{x-3}{x+2}\right) = \ln(x-3) - \ln(x+2) \]
Using the properties of logarithms, we can further simplify:
\[ \ln\left(\frac{x-3}{x+2}\right) = \ln\left(1 - \frac{3}{x}\right) - \ln\left(1 + \frac{2}{x}\right) \]

### Step 4: Use Taylor expansion for logarithm
For large \(x\), we can use the Taylor expansion:
\[ \ln(1 + u) \approx u \quad \text{for small } u \]
Thus,
\[ \ln\left(1 - \frac{3}{x}\right) \approx -\frac{3}{x} \quad \text{and} \quad \ln\left(1 + \frac{2}{x}\right) \approx \frac{2}{x} \]
So,
\[ \ln\left(\frac{x-3}{x+2}\right) \approx -\frac{3}{x} - \frac{2}{x} = -\frac{5}{x} \]

### Step 5: Substitute back into the limit
Now substitute this back into our limit expression:
\[ x \ln\left(\frac{x-3}{x+2}\right) \approx x \left(-\frac{5}{x}\right) = -5 \]
Thus, we have:
\[ \lim_{x \to \infty} x \ln\left(\frac{x-3}{x+2}\right) = -5 \]

### Step 6: Final limit calculation
Now substitute this result back into our exponent:
\[ \lim_{x \to \infty} e^{x \ln\left(\frac{x-3}{x+2}\right)} = e^{-5} \]

### Final Answer
\[ \lim_{x \to \infty} \left(\frac{x-3}{x+2}\right)^{x} = e^{-5} \]

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