Computing A Limit: Sin(x)/x As X Approaches Infinity
Let's dive into calculating a fascinating limit problem that often pops up in university admission textbooks. We're tackling this expression: $\lim_{x \to \infty}{{| \frac{sinx}{x}|}^{ \frac{sinx}{x}}}$. It looks intimidating at first, but don't worry, we'll break it down step by step to make it super clear.
Understanding the Problem
The first thing to notice is that we have a function raised to the power of itself, and that function involves . As x approaches infinity, the sine function oscillates between -1 and 1. However, it's being divided by x, which is growing without bound. This is a crucial observation because it helps us understand the behavior of the expression as x gets incredibly large. Essentially, we need to analyze how this fraction behaves and then figure out how that affects the overall limit.
The sine function, , is bounded between -1 and 1. This means that for any value of x, . When we divide by x, we are squeezing this bounded function by an increasingly large denominator. As x tends to infinity, the fraction approaches zero. This is a fundamental concept in calculus and is often referred to as the Squeeze Theorem or the Sandwich Theorem. The absolute value ensures that we are dealing with non-negative values, which is important when considering exponentiation. Thus, we are really looking at values approaching 0 from the positive side.
Initial Substitution and Challenges
So, the initial idea might be a simple substitution: let $t = \frac{sin(x)}{x}$. As $x \to \infty$, $t \to 0$. Then the limit becomes $\lim_{t \to 0} {|t|^t}$. But hold on, we need to be a bit careful here! The sine function can be both positive and negative, so $ \frac{sin(x)}{x}$ can approach zero from both positive and negative directions. However, the absolute value makes the base always positive. So the substitution approach is valid.
Why This Problem Is Interesting
This problem is a classic example of how limits can be deceptive. It involves a combination of trigonometric functions, rational functions, and exponentiation, forcing us to think critically about the behavior of each component as x tends to infinity. Moreover, the absolute value adds another layer of complexity, ensuring we're dealing with non-negative values. It's not just about blindly applying rules; it's about understanding the underlying principles.
Detailed Solution
Let's proceed with the substitution $t = \fracsin(x)}{x}$. As $x \rightarrow \infty$, $t \rightarrow 0$. Therefore, we want to find {|t|^t}$.
Since we are considering the absolute value, we can rewrite this as: $\lim_t \to 0^+} {t^t}$. To evaluate this limit, we can use the property $a^b = e^{b \ln a}$. So, we have$. Now, we need to find: $\lim_{t \to 0^+} {e^{t \ln t}} = e^{\lim_{t \to 0^+} {t \ln t}}$.
Now, let's find the limit of the exponent: $\lim_t \to 0^+} {t \ln t}$. This is an indeterminate form of the type $0 \cdot (-\infty)$. We can rewrite it as a fraction to apply L'Hôpital's Rule \frac{\ln t}{\frac{1}{t}}}$. This is now in the form $\frac{-\infty}{\infty}$, so we can apply L'Hôpital's Rule {\frac{\frac{1}{t}}{-\frac{1}{t^2}}} = \lim_{t \to 0^+} {(-t)} = 0$.
Therefore, the limit of the exponent is 0. Now, we can substitute this back into our original expression: $e^{\lim_{t \to 0^+} {t \ln t}} = e^0 = 1$.
Final Answer
Thus, $\lim_{x \to \infty}{{| \frac{sinx}{x}|}^{ \frac{sinx}{x}}} = 1$.
Key Takeaways
- Substitution: Recognizing the opportunity to simplify the problem by substituting $t = \frac{sin(x)}{x}$.
- L'Hôpital's Rule: Skillfully applying L'Hôpital's Rule to resolve the indeterminate form $0 \cdot (-\infty)$.
- Exponential Transformation: Using the property $a^b = e^{b \ln a}$ to transform the expression into a more manageable form.
- Understanding Limits: Grasping the concept of how $ \frac{sin(x)}{x}$ behaves as x approaches infinity.
Additional Insights
Squeeze Theorem
The Squeeze Theorem (also known as the Sandwich Theorem) is invaluable when dealing with limits involving trigonometric functions. It states that if we can "squeeze" a function between two other functions that have the same limit at a certain point, then the function in the middle must also have the same limit at that point. In this case, since $-1 \leq sin(x) \leq 1$, we have $-\frac{1}{x} \leq \frac{sin(x)}{x} \leq \frac{1}{x}$. As $x \to \infty$, both $-\frac{1}{x}$ and $\frac{1}{x}$ approach 0. Therefore, by the Squeeze Theorem, $\lim_{x \to \infty} \frac{sin(x)}{x} = 0$.
Indeterminate Forms
Recognizing and handling indeterminate forms is crucial in evaluating limits. Common indeterminate forms include $\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $,\infty - \infty$, $0^0$, $1^\infty$, and $,\infty^0$. L'Hôpital's Rule is often used to resolve indeterminate forms of the type $\frac{0}{0}$ and $\frac{\infty}{\infty}$. However, for other forms like $0 \cdot \infty$, algebraic manipulation is often required to transform them into a suitable form for L'Hôpital's Rule.
Importance of Absolute Value
The absolute value in the expression ${\left| \frac{sin(x)}{x} \right|}^{\frac{sin(x)}{x}}$ ensures that the base is always non-negative. This is important because the function $t^t$ is not defined for negative values of t. By taking the absolute value, we ensure that we are only dealing with positive values, which allows us to apply the necessary techniques to evaluate the limit. Without the absolute value, the problem would become significantly more complex, as we would need to consider the cases where $\frac{sin(x)}{x}$ is positive and negative separately.
Conclusion
So, there you have it! By using substitution, L'Hôpital's Rule, and a solid understanding of limits, we've successfully computed $\lim_{x \to \infty}{{| \frac{sinx}{x}|}^{ \frac{sinx}{x}}} = 1$. Keep practicing, and you'll become a limit-solving pro in no time!