Question

find the limits as $x \to \infty$ and as $x \to -\infty$. (if the limi $f(x) = (3 - x)(1 + x)^2(1 - x)^4$ $\lim\limits_{x \to \infty} f(x) = \square$ $\lim\limits_{x \to -\infty} f(x) = \square$

Explanation:

Step1: Determine the leading term

To find the limit as \( x \to \pm\infty \), we focus on the leading term of the polynomial. First, expand the powers of each factor:

• For \( (3 - x) \), the leading term is \( -x \).
• For \( (1 + x)^2 \), expanding gives \( x^2 + 2x + 1 \), so the leading term is \( x^2 \).
• For \( (1 - x)^4 \), we can rewrite it as \( (-(x - 1))^4 = (x - 1)^4 \). Expanding \( (x - 1)^4 \) (using the binomial theorem or recognizing the leading term), the leading term is \( x^4 \).

Now, multiply the leading terms together:
\( (-x) \cdot (x^2) \cdot (x^4) = -x^{1 + 2 + 4} = -x^7 \).

Step2: Find \( \lim_{x \to \infty} f(x) \)

As \( x \to \infty \), we analyze the leading term \( -x^7 \). When \( x \) becomes very large (positive), \( x^7 \) is positive, and multiplying by \( -1 \) gives a very large negative number. So:
\( \lim_{x \to \infty} f(x) = \lim_{x \to \infty} -x^7 = -\infty \).

Step3: Find \( \lim_{x \to -\infty} f(x) \)

As \( x \to -\infty \), we look at the leading term \( -x^7 \). When \( x \) is a large negative number, \( x^7 \) is negative (since the exponent 7 is odd: \( (-a)^7 = -a^7 \) for \( a > 0 \)). Then \( -x^7 = -(-a^7) = a^7 \), which is a large positive number. So:
\( \lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} -x^7 = \infty \).

Answer:

\( \lim_{x \to \infty} f(x) = \boxed{-\infty} \)
\( \lim_{x \to -\infty} f(x) = \boxed{\infty} \)