Evaluate the following limits :
`Lim_(x to oo) ((x+1)(2x+3))/((x+2)(3x+4))`

Answer

To evaluate the limit

\[ \lim_{x \to \infty} \frac{(x+1)(2x+3)}{(x+2)(3x+4)}, \]

we will follow these steps:

### Step 1: Identify the limit form
As \( x \) approaches infinity, both the numerator and denominator approach infinity. Thus, we have an indeterminate form of type \(\frac{\infty}{\infty}\).

### Step 2: Factor out the highest power of \( x \)
In both the numerator and denominator, we will factor out \( x \) from each term:

\[ \text{Numerator: } (x+1)(2x+3) = x(1 + \frac{1}{x}) \cdot x(2 + \frac{3}{x}) = x^2(1 + \frac{1}{x})(2 + \frac{3}{x}). \]

\[ \text{Denominator: } (x+2)(3x+4) = x(1 + \frac{2}{x}) \cdot x(3 + \frac{4}{x}) = x^2(1 + \frac{2}{x})(3 + \frac{4}{x}). \]

### Step 3: Rewrite the limit
Now we can rewrite the limit as:

\[ \lim_{x \to \infty} \frac{x^2(1 + \frac{1}{x})(2 + \frac{3}{x})}{x^2(1 + \frac{2}{x})(3 + \frac{4}{x})}. \]

### Step 4: Cancel \( x^2 \)
Since \( x^2 \) appears in both the numerator and denominator, we can cancel it out:

\[ \lim_{x \to \infty} \frac{(1 + \frac{1}{x})(2 + \frac{3}{x})}{(1 + \frac{2}{x})(3 + \frac{4}{x})}. \]

### Step 5: Evaluate the limit as \( x \to \infty \)
Now we can substitute \( x \to \infty \):

- As \( x \to \infty \), \( \frac{1}{x} \to 0 \).
- Thus, we have:

\[ \lim_{x \to \infty} \frac{(1 + 0)(2 + 0)}{(1 + 0)(3 + 0)} = \frac{1 \cdot 2}{1 \cdot 3} = \frac{2}{3}. \]

### Final Answer
The limit is

\[ \frac{2}{3}. \]

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