`lim_(x to infty) ((2+x)^(40) (4+x)^(5))/(2-x)^(45)` =

Options

A. `-1`
B. `1
C. 16
D. 32

Answer

To solve the limit \( \lim_{x \to \infty} \frac{(2+x)^{40} (4+x)^{5}}{(2-x)^{45}} \), we can follow these steps:

### Step 1: Simplify the expression
We start by rewriting the terms in the limit to make it easier to analyze as \( x \) approaches infinity.

\[ \lim_{x \to \infty} \frac{(2+x)^{40} (4+x)^{5}}{(2-x)^{45}} = \lim_{x \to \infty} \frac{(x(1+\frac{2}{x}))^{40} (x(1+\frac{4}{x}))^{5}}{(x(-1+\frac{2}{x}))^{45}} \]

### Step 2: Factor out \( x \)
Now we factor out \( x \) from each term:

\[ = \lim_{x \to \infty} \frac{x^{40} (1+\frac{2}{x})^{40} \cdot x^{5} (1+\frac{4}{x})^{5}}{x^{45} (-1+\frac{2}{x})^{45}} \]

### Step 3: Combine the powers of \( x \)
Combine the powers of \( x \):

\[ = \lim_{x \to \infty} \frac{x^{45} (1+\frac{2}{x})^{40} (1+\frac{4}{x})^{5}}{x^{45} (-1+\frac{2}{x})^{45}} \]

### Step 4: Cancel \( x^{45} \)
We can cancel \( x^{45} \) from the numerator and denominator:

\[ = \lim_{x \to \infty} \frac{(1+\frac{2}{x})^{40} (1+\frac{4}{x})^{5}}{(-1+\frac{2}{x})^{45}} \]

### Step 5: Evaluate the limit
As \( x \) approaches infinity, \( \frac{2}{x} \) and \( \frac{4}{x} \) both approach 0. Thus, we can evaluate the limit:

\[ = \frac{(1+0)^{40} (1+0)^{5}}{(-1+0)^{45}} = \frac{1 \cdot 1}{(-1)^{45}} = \frac{1}{-1} = -1 \]

### Final Result
Thus, the limit is:

\[ \lim_{x \to \infty} \frac{(2+x)^{40} (4+x)^{5}}{(2-x)^{45}} = -1 \]

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