Question:medium

Evaluate: \[ \lim_{x \to \infty} \left( \frac{x+8}{x+1} \right)^{x+5} = \, ? \] 

Show Hint

Whenever you see an expression of the form \[ \left(\frac{x+a}{x+b}\right)^x \] first rewrite the base as \[ 1+\frac{a-b}{x+b} \] and then use the standard exponential limit.
Updated On: May 14, 2026
  • \(e^4\)
  • \(e^5\)
  • \(e^{11}\)
  • \(e^7\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This limit is of the indeterminant form \(1^\infty\), as the base tends to 1 and the exponent tends to infinity.
Step 2: Key Formula or Approach:
For \(\lim_{x \to a} [f(x)]^{g(x)}\) where \(f(x) \to 1\) and \(g(x) \to \infty\):
\[ \text{Limit} = e^{\lim_{x \to a} g(x)[f(x) - 1]} \] Step 3: Detailed Explanation:
Identify the functions: \(f(x) = \frac{x+8}{x+1}\) and \(g(x) = x+5\).
Apply the formula:
\[ \text{Limit} = e^{\lim_{x \to \infty} (x+5) [ \frac{x+8}{x+1} - 1 ]} \] Simplify the term in brackets:
\[ \frac{x+8}{x+1} - 1 = \frac{x+8-(x+1)}{x+1} = \frac{7}{x+1} \] Now evaluate the limit in the exponent:
\[ \lim_{x \to \infty} \frac{7(x+5)}{x+1} = \lim_{x \to \infty} \frac{7x+35}{x+1} \] Using the rule for limits at infinity for rational functions:
\[ \lim_{x \to \infty} \frac{7 + 35/x}{1 + 1/x} = 7 \] Thus, the final limit is \(e^7\).
Step 4: Final Answer:
The value of the limit is \(e^7\).
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