Question:medium

The value of $\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4}$ is:

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For $\lim_{x \to \infty} \left(\frac{x+a}{x+b}\right)^{x+c}$, the limit is always $e^{a-b}$. Here, $a=6, b=1$, so the answer is $e^{6-1} = e^5$. This shortcut takes only 2 seconds!
Updated On: May 31, 2026
  • $e^5$
  • $e^6$
  • $e$
  • $e^4$
Show Solution

The Correct Option is A

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