如何将无穷限积分化简的∞-∞型极限转为∞/∞型以应用洛必达法则？

Absolutely! Converting that ∞−∞ indeterminate form into ∞/∞ is totally feasible here, and it’s actually the standard go-to move for handling these logarithmic terms. Let me walk you through the process step by step:

Step 1: Merge the logarithmic terms using logarithm rules #

First, factor out the smallest common coefficient from the two ln terms to combine them into a single logarithm (remember that (a\ln b - c\ln d = \ln\left(\frac{b^a}{dc}\right))):
[
\frac{\ln(t+5)}{26} - \frac{\ln(t^2+1)}{52} = \frac{1}{52}\left[2\ln(t+5) - \ln(t^2+1)\right] = \frac{1}{52}\ln\left(\frac{(t+5)^2}{t2+1}\right)
]

Step 2: Expose the ∞/∞ form (and simplify if possible) #

Now your entire limit expression becomes:
[
\lim_{t \to \infty} \left[ \frac{1}{52}\ln\left(\frac{(t+5)^2}{t2+1}\right) + \frac{5\arctan(t)}{26} - \frac{\ln5}{26} \right]
]
The key part here is the rational function inside the logarithm: (\frac{(t+5)^2}{t2+1}) is an ∞/∞ indeterminate form as (t \to \infty). You can apply L’Hospital’s Rule here if you want, but a quicker approach is to look at the leading terms:

• Expanding the numerator: ((t+5)^2 = t^2 + 10t + 25)
• Denominator: (t^2 + 1)
  As (t) blows up to infinity, the lower-degree terms (10t, 25, 1) become irrelevant. So (\frac{(t+5)^2}{t2+1} \to \frac{t^2}{t2} = 1), which means (\ln(1) = 0). That entire logarithmic term drops out.

Step 3: Evaluate the remaining straightforward terms #

• For the arctangent term: We know (\lim_{t \to \infty} \arctan(t) = \frac{\pi}{2}), so (\frac{5\arctan(t)}{26}) approaches (\frac{5\pi}{52}).
• The constant term (-\frac{\ln5}{26}) doesn’t change as (t) goes to infinity.

Putting everything together, the final value of the limit is (\frac{5\pi}{52} - \frac{\ln5}{26}).

Quick note on why this conversion works #

∞−∞ forms involving logarithms almost always benefit from combining the logs into a single log of a rational function. This naturally transforms the problematic ∞−∞ into an ∞/∞ form that’s easy to handle (either with L’Hospital’s Rule or leading-term analysis, which is faster here). Trying to apply L’Hospital’s directly to the original ∞−∞ expression would be messy and unnecessary—this method is way cleaner.

内容的提问来源于stack exchange，提问作者Art