Question

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

• A. $e^5$
• B. $e^6$
• C. $e$
• D. $e^4$

Hint:

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!

Answer 1

The Correct Option is A

Step 1: Spot the form of the limit.
As $x$ grows huge, the fraction $\frac{x+6}{x+1}$ gets very close to $1$, while the power $x+4$ gets very large. So this is the special $1^\infty$ case, which needs a trick, not direct substitution.

Step 2: Recall the handy formula.
When the base goes to $1$ and the power goes to infinity, the answer is $e$ raised to a special limit.
\[ \lim [f(x)]^{g(x)} = e^{\lim (f(x) - 1)\,g(x)} \] Here $f(x) = \frac{x+6}{x+1}$ and $g(x) = x+4$.

Step 3: Find $f(x) - 1$.
Subtract $1$ from the fraction by making a common bottom.
\[ \frac{x+6}{x+1} - 1 = \frac{(x+6) - (x+1)}{x+1} = \frac{5}{x+1} \]

Step 4: Multiply by $g(x)$.
Now multiply this by the power $x+4$.
\[ \frac{5}{x+1}\times (x+4) = \frac{5x + 20}{x+1} \]

Step 5: Take the limit of this part.
For large $x$, the top and bottom are both close to $5x$ and $x$. Dividing the leading terms gives $5$.
\[ \lim_{x\to\infty} \frac{5x+20}{x+1} = 5 \]

Step 6: Raise $e$ to this value.
The whole limit is $e$ to the power we just found.
\[ e^5 \]
\[ \boxed{e^5} \]