Question:medium

\[ \lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x= \]

Show Hint

Always remember the standard limit \(\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e\). It is used frequently in limits.
  • \(0\)
  • \(1\)
  • \(e\)
  • \(\infty\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
This question involves evaluating a fundamental limit in calculus that arises from the study of continuous compounding and growth.
The expression is of the form \( 1^\infty \), which is an indeterminate form.
This particular limit is used to define Euler's number, denoted as \( e \).
Step 2: Key Formula or Approach:
1. If a limit is of the form \( \lim_{x \to a} [f(x)]^{g(x)} \) where \( f(x) \to 1 \) and \( g(x) \to \infty \), it can be evaluated as \( e^{\lim_{x \to a} (f(x) - 1) g(x)} \).
2. Alternatively, recognize this as the standard definition of the exponential constant \( e \).
Step 3: Detailed Explanation:

Let the limit be \( L = \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x \).

As \( x \) approaches infinity, \( 1/x \) approaches 0. Thus the base \( (1 + 1/x) \) approaches 1.

The exponent \( x \) approaches infinity. This gives us the \( 1^\infty \) form.

Let's use the natural logarithm to simplify:
\[ \ln L = \lim_{x \to \infty} x \ln\left(1 + \frac{1}{x}\right) \]

Rewrite as:
\[ \ln L = \lim_{x \to \infty} \frac{\ln(1 + 1/x)}{1/x} \]

Let \( t = 1/x \). As \( x \to \infty \), \( t \to 0 \).
\[ \ln L = \lim_{t \to 0} \frac{\ln(1 + t)}{t} \]

Using the standard limit property \( \lim_{t \to 0} \frac{\ln(1 + t)}{t} = 1 \):
\[ \ln L = 1 \]

Taking the exponential of both sides:
\[ L = e^1 = e \]

Therefore, the limiting value of the growth sequence is the irrational constant \( e \approx 2.71828 \).

Step 4: Final Answer:
The value of the limit is \( e \).
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