求极限问题咨询：两道x→∞时的极限求解困惑

Hey there, let’s tackle these two limits one by one—hyperbolic functions can feel tricky at infinity, but unpacking them using their exponential definitions will make this straightforward.

First, recall the exponential definition of hyperbolic sine:
$$\sinh(t) = \frac{e^t - e^{-t}}{2}$$

Substitute $t = x^2$ into the expression:
$$\log \sinh(x^2) = \log\left( \frac{e^{x2} - e^{-x2}}{2} \right)$$

Factor out $e^{x2}$ from the numerator (it’s the dominant term as $x \to \infty$):
$$\log\left( e^{x2} \cdot \frac{1 - e^{-2x2}}{2} \right)$$

Use logarithm properties to split this into simpler terms:
$$\log(e^{x2}) + \log\left( \frac{1 - e^{-2x2}}{2} \right) = x^2 + \log(1 - e^{-2x2}) - \log 2$$

Now substitute back into the original expression:
$$\left( x^2 + \log(1 - e^{-2x2}) - \log 2 \right) - x^2 = \log(1 - e^{-2x2}) - \log 2$$

As $x \to \infty$, $e^{-2x2}$ approaches 0 (since the exponent plummets to $-\infty$). So $\log(1 - 0) = 0$, leaving us with:
$$0 - \log 2 = -\log 2 \quad (\text{or equivalently } \log\left( \frac{1}{2} \right))$$

Start with the exponential definition of hyperbolic tangent:
$$\tanh(x) = \frac{e^x - e^{-x}}{ex + e^{-x}}$$

First simplify the numerator $\tanh(x) - 1$ by combining terms over a common denominator:
$$\frac{e^x - e^{-x}}{ex + e^{-x}} - 1 = \frac{e^x - e^{-x} - (e^x + e^{-x})}{ex + e^{-x}} = \frac{-2e^{-x}}{ex + e^{-x}}$$

Now substitute this back into the original fraction:
$$\frac{\frac{-2e^{-x}}{ex + e^{-x}}}{e{-2x}} = \frac{-2e^{-x}}{e{-2x}(e^x + e^{-x})}$$

Simplify the denominator by distributing $e^{-2x}$:
$$\frac{-2e^{-x}}{e{-x} + e^{-3x}}$$

Divide both numerator and denominator by $e^{-x}$ (the dominant term in the denominator as $x \to \infty$):
$$\frac{-2}{1 + e^{-2x}}$$

As $x \to \infty$, $e^{-2x}$ approaches 0, so the limit becomes:
$$\frac{-2}{1 + 0} = -2$$

Hope that clears things up! The key trick here is always going back to the exponential roots of hyperbolic functions—they make the behavior at infinity much easier to track.

内容的提问来源于stack exchange，提问作者Niktaneous