Mastering The Limit $7 Lim_{x→0} X / Sin Bx$ A Comprehensive Guide

Hey guys! Today, we're tackling a classic calculus problem that often pops up in introductory courses: evaluating the limit $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx}$. This limit is a fantastic example of how understanding fundamental trigonometric limits and limit laws can help us solve more complex problems. We'll break down the problem step-by-step, making sure you grasp the underlying concepts and can confidently tackle similar challenges in the future. So, grab your thinking caps, and let's dive in!

Understanding the Foundation: The Limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$

Before we can conquer our main problem, we need to understand a crucial cornerstone of trigonometric limits: the limit of $\frac{\sin x}{x}$ as $x$ approaches 0. This limit is not immediately obvious. If we directly substitute $x = 0$, we get the indeterminate form $\frac{0}{0}$. This means we can't just plug in the value; we need to use some mathematical trickery to find the true limit. The standard way to prove this limit is to use the Squeeze Theorem (also known as the Sandwich Theorem). This theorem states that if we can "squeeze" a function between two other functions that both approach the same limit, then our function must also approach that limit. To prove $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$, we typically use geometric arguments involving the unit circle and comparing areas of triangles and sectors. Imagine a unit circle (a circle with radius 1) and an angle $x$ (in radians) in the first quadrant. We can construct a triangle with vertices at the origin, the point where the angle intersects the circle, and a point on the x-axis. We can also construct a sector of the circle subtended by the angle $x$, and another triangle formed by extending the side of the angle to intersect a vertical line at $x = 1$. By comparing the areas of these three figures (triangle inside the sector, sector inside the larger triangle), we can establish the following inequality: $\sin x < x < \tan x$ for $0 < x < \frac{\pi}{2}$. Dividing all parts of the inequality by $\sin x$ (which is positive in the first quadrant), we get $1 < \frac{x}{\sin x} < \frac{1}{\cos x}$. Taking the reciprocal of each part (and reversing the inequality signs), we have $\cos x < \frac{\sin x}{x} < 1$. Now, as $x$ approaches 0, $\cos x$ approaches 1. Therefore, we have $\lim_{x \rightarrow 0} \cos x = 1$ and $\lim_{x \rightarrow 0} 1 = 1$. By the Squeeze Theorem, since $\frac{\sin x}{x}$ is squeezed between $\cos x$ and 1, we must have $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$. This limit is a fundamental result in calculus, and it's worth memorizing and understanding its proof. It's the key that unlocks the solution to our original problem!

Tackling Our Main Problem: $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx}$

Okay, now that we have the fundamental limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ under our belts, let's attack our main problem: $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx}$. Remember, the goal is to manipulate the expression inside the limit so that it resembles the fundamental limit we just learned. The first thing we can do is deal with the constant 7. A crucial property of limits is that the limit of a constant times a function is the constant times the limit of the function. Mathematically, this is expressed as $\lim_{x \rightarrow a} [cf(x)] = c \lim_{x \rightarrow a} f(x)$, where 'c' is a constant. So, we can pull the 7 outside the limit: $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx} = 7 \cdot \lim_{x \rightarrow 0} \frac{x}{\sin bx}$. Now, let's focus on the expression $\frac{x}{\sin bx}$. We want to make the denominator look like $\frac{\sin u}{u}$ for some expression 'u' that approaches 0 as $x$ approaches 0. In our case, the argument of the sine function is $bx$. So, we want the denominator to look like $\sin bx / bx$. To achieve this, we can multiply the numerator and denominator by 'b': $7 \cdot \lim_{x \rightarrow 0} \frac{x}{\sin bx} = 7 \cdot \lim_{x \rightarrow 0} \frac{x}{\sin bx} \cdot \frac{b}{b} = 7 \cdot \lim_{x \rightarrow 0} \frac{bx}{b \sin bx}$. Now we can rearrange the terms: $7 \cdot \lim_{x \rightarrow 0} \frac{bx}{b \sin bx} = 7 \cdot \lim_{x \rightarrow 0} \frac{1}{b} \cdot \frac{bx}{\sin bx}$. Again, we can pull the constant $\frac{1}{b}$ outside the limit (assuming $b$ is not zero, otherwise the original expression is undefined): $7 \cdot \lim_{x \rightarrow 0} \frac{1}{b} \cdot \frac{bx}{\sin bx} = \frac{7}{b} \cdot \lim_{x \rightarrow 0} \frac{bx}{\sin bx}$. Almost there! We have $\frac{bx}{\sin bx}$, which is the reciprocal of the fundamental limit we know. Remember, $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$, and if a limit is 1, its reciprocal also has a limit of 1. So, $\lim_{x \rightarrow 0} \frac{x}{\sin x} = 1$. Now, let's consider the limit $\lim_{x \rightarrow 0} \frac{bx}{\sin bx}$. As $x$ approaches 0, $bx$ also approaches 0. So, we can make a substitution: let $u = bx$. As $x \rightarrow 0$, $u \rightarrow 0$. Then, $\lim_{x \rightarrow 0} \frac{bx}{\sin bx} = \lim_{u \rightarrow 0} \frac{u}{\sin u}$. Since we know that $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$, we also know that $\lim_{u \rightarrow 0} \frac{\sin u}{u} = 1$. Therefore, $\lim_{u \rightarrow 0} \frac{u}{\sin u} = \frac{1}{\lim_{u \rightarrow 0} \frac{\sin u}{u}} = \frac{1}{1} = 1$. Putting it all together, we have: $\frac{7}{b} \cdot \lim_{x \rightarrow 0} \frac{bx}{\sin bx} = \frac{7}{b} \cdot 1 = \frac{7}{b}$. So, the final answer is $\frac{7}{b}$.

Key Takeaways and General Strategies

Wow, we did it! We successfully evaluated the limit $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx}$. Let's recap the key steps and strategies we used, as these are applicable to many other limit problems you'll encounter.

• Master the Fundamental Limits: The limit $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$ is a cornerstone of trigonometric limits. Make sure you understand it, memorize it, and know how to prove it (using the Squeeze Theorem). This limit (and its reciprocal) frequently appears in more complex problems.
• Recognize Indeterminate Forms: When evaluating limits, the first step is always to try direct substitution. If you get an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, it means you need to do more work. These forms indicate that the limit might exist, but you need to manipulate the expression to find it.
• Manipulate the Expression: The key to solving many limit problems is to manipulate the expression so that it resembles a known limit. This often involves algebraic techniques like multiplying by a clever form of 1, factoring, rationalizing, or using trigonometric identities. In our problem, we multiplied the numerator and denominator by 'b' to make the denominator look like $\frac{\sin bx}{bx}$.
• Use Limit Laws: Limit laws are your friends! They allow you to break down complex limits into simpler ones. We used the constant multiple rule ($\lim_{x \rightarrow a} [cf(x)] = c \lim_{x \rightarrow a} f(x)$) and the reciprocal rule (if $\lim_{x \rightarrow a} f(x) = L$ and $L \neq 0$, then $\lim_{x \rightarrow a} \frac{1}{f(x)} = \frac{1}{L}$). Understanding and applying these laws simplifies the process significantly.
• Substitution: Sometimes, a well-chosen substitution can make a limit problem much easier. In our case, substituting $u = bx$ allowed us to directly apply the fundamental limit $\lim_{u \rightarrow 0} \frac{\sin u}{u} = 1$.

By understanding these concepts and practicing these strategies, you'll be well-equipped to tackle a wide range of limit problems. Remember, calculus is all about building on fundamental ideas, so mastering these basics is essential for your success.

Practice Makes Perfect: Example Problems

To solidify your understanding, let's look at a couple of similar problems that you can try on your own. Working through these will help you internalize the techniques we've discussed and build your confidence.

Example 1: Evaluate $\lim_{x \rightarrow 0} \frac{\sin 3x}{x}$.

Hint: Think about how you can manipulate the denominator to involve $3x$.

Example 2: Evaluate $\lim_{x \rightarrow 0} \frac{\tan x}{x}$.

Hint: Recall that $\tan x = \frac{\sin x}{\cos x}$. Can you rewrite the expression and use the fundamental limit?

Try these problems out! Working through them step-by-step, using the strategies we've discussed, will help you truly master these concepts. If you get stuck, don't worry! Review the steps we took in the original problem, and remember the key ideas. Calculus is a journey, and every problem you solve brings you closer to mastery.

Conclusion: You've Got This!

So, there you have it! We've successfully navigated the limit $7 \lim_{x \rightarrow 0} \frac{x}{\sin bx}$, broken down the underlying concepts, and explored the strategies you can use to tackle similar problems. Remember, the key is to understand the fundamental limits, manipulate expressions to match those limits, and use limit laws to your advantage. With practice and persistence, you'll become a limit-solving pro in no time. Keep practicing, keep exploring, and keep asking questions. You've got this!