含余弦、正弦函数的负无穷极限求解咨询

Hey there! Let's break down this limit problem step by step—those cosine and sine functions can seem intimidating, but their boundedness is actually our secret weapon here. Let's start with the expression you're working on:

$$\lim_{x\to-\infty} \frac{x\cos(x^{{2})+3x\sqrt{1-4x}}{\sqrt[4]{x}{6}-x^{5}}+\sin(x{5})}$$

Key Observations First #

• The functions $\cos(x^2)$ and $\sin(x^5)$ are bounded: for all real $x$, $|\cos(x^2)| \leq 1$ and $|\sin(x^5)| \leq 1$. When multiplied by terms that tend to 0 or are negligible compared to dominant terms, these will disappear in the limit.
• When $x \to -\infty$, the highest-power terms in radicals and polynomials will dominate—we can focus on those and ignore smaller, lower-power terms.

Step 1: Simplify the Dominant Terms #

Let's first handle the numerator and denominator separately, focusing on their largest terms:

1. Numerator: The two terms are $x\cos(x^2)$ and $3x\sqrt{1-4x}$.

   • $x\cos(x^2)$ is $x$ multiplied by a bounded function. As $x \to -\infty$, we'll see this term is negligible compared to the second one.
   • For $3x\sqrt{1-4x}$, when $x \to -\infty$, $-4x$ dominates inside the square root. So:
     $$\sqrt{1-4x} = \sqrt{-4x\left(1 - \frac{1}{4x}\right)} = 2\sqrt{-x} \cdot \sqrt{1 - \frac{1}{4x}}$$
     (Note: $\sqrt{-x}$ is real because $x$ is negative here.)
     Substituting back, the second term becomes:
     $$3x \cdot 2\sqrt{-x} \cdot \sqrt{1 - \frac{1}{4x}} = 6x\sqrt{-x} \cdot \sqrt{1 - \frac{1}{4x}}$$
     Since $x = -\sqrt{-x} \cdot \sqrt{-x}$, we can rewrite $x\sqrt{-x} = -(-x)^{3/2}$, so this term simplifies to:
     $$-6(-x)^{3/2} \cdot \sqrt{1 - \frac{1}{4x}}$$

2. Denominator: The two terms are $\sqrt[4]{x^6 - x^5}$ and $\sin(x^5)$.

   • $\sin(x^5)$ is bounded, so it's negligible compared to the radical term (which tends to $+\infty$ as $x \to -\infty$).
   • For $\sqrt[4]{x^6 - x^5}$, factor out $x^6$ (the dominant term):
     $$\sqrt[4]{x^6\left(1 - \frac{1}{x}\right)} = |x|^{6/4} \cdot \sqrt[4]{1 - \frac{1}{x}}$$
     Since $x \to -\infty$, $|x| = -x$, so $|x|^{3/2} = (-x)^{3/2}$. Thus the radical becomes:
     $$(-x)^{3/2} \cdot \sqrt[4]{1 - \frac{1}{x}}$$

Step 2: Cancel Dominant Terms and Evaluate the Limit #

Now, substitute the simplified dominant terms back into the limit, ignoring the negligible bounded terms:
$$\lim_{x\to-\infty} \frac{-6(-x)^{3/2} \cdot \sqrt{1 - \frac{1}{4x}}}{(-x)^{3/2} \cdot \sqrt[4]{1 - \frac{1}{x}}}$$

The $(-x)^{3/2}$ terms cancel out. As $x \to -\infty$, $\frac{1}{4x}$ and $\frac{1}{x}$ both tend to 0, so $\sqrt{1 - 0} = 1$ and $\sqrt[4]{1 - 0} = 1$. This leaves us with:
$$\lim_{x\to-\infty} -6 \cdot \frac{1}{1} = -6$$

Why We Can Ignore the Bounded Terms #

• For the numerator's $x\cos(x^2)$ term: when we divide by $(-x)^{3/2}$, we get $\frac{x\cos(x^2)}{(-x){3/2}} = \frac{x}{-x\sqrt{-x}} \cdot \cos(x^2) = -\frac{\cos(x^2)}{\sqrt{-x}}$. Since $\cos(x^2)$ is bounded and $\sqrt{-x} \to +\infty$, this tends to 0.
• For the denominator's $\sin(x^5)$ term: dividing by $(-x)^{3/2}$ gives $\frac{\sin(x^5)}{(-x){3/2}}$, which also tends to 0 because the numerator is bounded and the denominator tends to $+\infty$.

So putting it all together, the limit is -6.

备注：内容来源于stack exchange，提问作者Ronan Finn