Question:
The value of $\lim_{x \to \infty} \left(\frac{x+6}{x+1}\right)^{x+4}$ is:
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: Jun 3, 2026
- $e^5$
- $e^6$
- $e$
- $e^4$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
For limits of the indeterminate form $1^\infty$, we use the standard theorem: if $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$, then: \[ \lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} (f(x) - 1)g(x)} \]
Step 2: Meaning
Here, $f(x) = \frac{x+6}{x+1}$ and $g(x) = x+4$. As $x \to \infty$, $f(x) \to 1$ and $g(x) \to \infty$.
Step 3: Analysis
Evaluate the exponent limit: \[ L = \lim_{x \to \infty} (f(x) - 1)g(x) = \lim_{x \to \infty} \left(\frac{x+6}{x+1} - 1\right)(x+4) \] \[ L = \lim_{x \to \infty} \left(\frac{x+6 - (x+1)}{x+1}\right)(x+4) = \lim_{x \to \infty} \left(\frac{5}{x+1}\right)(x+4) \] \[ L = \lim_{x \to \infty} \frac{5x + 20}{x+1} = 5 \] Thus, the limit value is $e^5$.
Step 4: Conclusion
The value of the limit is $e^5$.
Final Answer: (A)
For limits of the indeterminate form $1^\infty$, we use the standard theorem: if $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$, then: \[ \lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} (f(x) - 1)g(x)} \]
Step 2: Meaning
Here, $f(x) = \frac{x+6}{x+1}$ and $g(x) = x+4$. As $x \to \infty$, $f(x) \to 1$ and $g(x) \to \infty$.
Step 3: Analysis
Evaluate the exponent limit: \[ L = \lim_{x \to \infty} (f(x) - 1)g(x) = \lim_{x \to \infty} \left(\frac{x+6}{x+1} - 1\right)(x+4) \] \[ L = \lim_{x \to \infty} \left(\frac{x+6 - (x+1)}{x+1}\right)(x+4) = \lim_{x \to \infty} \left(\frac{5}{x+1}\right)(x+4) \] \[ L = \lim_{x \to \infty} \frac{5x + 20}{x+1} = 5 \] Thus, the limit value is $e^5$.
Step 4: Conclusion
The value of the limit is $e^5$.
Final Answer: (A)
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