Problem 201
Question
$$ \lim _{x \rightarrow 0} \frac{\sin 4 x}{1-\sqrt{1-x}} \cdot\\{\text { Ans. } 8\\} $$
Step-by-Step Solution
Verified Answer
The limit of the function \( \lim_{x \rightarrow 0} \frac{\sin 4x}{1-\sqrt{1-x}} \) is -8.
1Step 1: Identify the original form of the function.
Identify the original form of \(x\) in that function. Substituting \(x = 0\) into \(\frac{\sin 4 x}{1-\sqrt{1-x}}\), we get \(\frac{0}{1 - 1}\), a 0/0 form, which is undefined. So, L'Hopital's rule could be applied to solve this problem.
2Step 2: Apply the L'Hopital's rule
Take the derivatives of numerator and denominator. The derivative of \(\sin 4x\) with respect to \(x\) is \(4\cos 4x\). Using rules of derivatives, the derivative of \(1-\sqrt{1-x}\) with respect to \(x\) is \(-\frac{1}{2\sqrt{1-x}}\). Now, the function becomes \(\lim_{x \rightarrow 0} \frac{4\cos 4x}{-\frac{1}{2\sqrt{1-x}}}\). After simplifying the function, it becomes \(\lim_{x \rightarrow 0} -8\cos 4x\sqrt{1-x}\).
3Step 3: Substitute the x-value into the new function
Substitute \(x = 0\) into \(-8\cos 4x\sqrt{1-x}\). We get \( -8 \times \cos (4 \times 0) \times \sqrt{1 - 0} = -8\).
Key Concepts
Limits and ContinuityTrigonometric FunctionsDerivatives
Limits and Continuity
When evaluating limits in calculus, we are essentially determining what value a function approaches as the input gets close to a certain number. This process is critical for understanding the behavior of functions at points where they might not be well-defined. For instance, in the given problem, as \(x\) approaches 0, the original function \(\frac{\sin 4x}{1-\sqrt{1-x}}\) takes on an indeterminate form \(\frac{0}{0}\).
This tells us that evaluating these directly doesn't yield any meaningful result. That's where L'Hôpital's Rule becomes handy, allowing us to compute a workable limit by addressing this indeterminate form through differentiation.
Continuity plays a role here too. A function is continuous at a point if the limit as you approach that point is equal to the function's value at that point. Understanding limits helps assess and ensure a function's continuity at various points.
This tells us that evaluating these directly doesn't yield any meaningful result. That's where L'Hôpital's Rule becomes handy, allowing us to compute a workable limit by addressing this indeterminate form through differentiation.
Continuity plays a role here too. A function is continuous at a point if the limit as you approach that point is equal to the function's value at that point. Understanding limits helps assess and ensure a function's continuity at various points.
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in calculus. In our example, \(\sin 4x\) is used in the numerator of the original limit expression.
These functions oscillate between -1 and 1 and are periodic, meaning their values repeat in regular intervals. The derivative of a trigonometric function is a key concept, as it helps us understand the rate at which these functions change.
When applying L'Hôpital's rule, the derivative of \(\sin 4x\) becomes \(4\cos 4x\). This derivation reflects the chain rule, an essential rule when differentiating composite functions. The 4 comes from the inner function's derivative, \(4x\), making it a product with \(\cos 4x\). It's critical to grasp these derivatives as they are often used to simplify and evaluate limits effectively.
These functions oscillate between -1 and 1 and are periodic, meaning their values repeat in regular intervals. The derivative of a trigonometric function is a key concept, as it helps us understand the rate at which these functions change.
When applying L'Hôpital's rule, the derivative of \(\sin 4x\) becomes \(4\cos 4x\). This derivation reflects the chain rule, an essential rule when differentiating composite functions. The 4 comes from the inner function's derivative, \(4x\), making it a product with \(\cos 4x\). It's critical to grasp these derivatives as they are often used to simplify and evaluate limits effectively.
Derivatives
Derivatives are at the heart of calculus, representing the rate at which a quantity changes. In tackling the limit problem using L'Hôpital's Rule, finding the derivative of both the numerator and denominator is necessary.
For the numerator, \(\sin 4x\), the derivative is \(4\cos 4x\), as explained previously. For the denominator, \(1 - \sqrt{1-x}\), the derivative is \(-\frac{1}{2\sqrt{1-x}}\).
Derivatives unveil how functions behave. They skillfully turn complex calculations into simpler forms to find limits or solve dynamic problems. Understanding these basic derivative computations enables students to solve limits even when they appear difficult at first.
For the numerator, \(\sin 4x\), the derivative is \(4\cos 4x\), as explained previously. For the denominator, \(1 - \sqrt{1-x}\), the derivative is \(-\frac{1}{2\sqrt{1-x}}\).
Derivatives unveil how functions behave. They skillfully turn complex calculations into simpler forms to find limits or solve dynamic problems. Understanding these basic derivative computations enables students to solve limits even when they appear difficult at first.
Other exercises in this chapter
Problem 199
$$ \lim _{x \rightarrow 0} \frac{x\left(5^{x}-1\right)}{1-\cos x} .\\{\text { Ans. } 2 \ln 5\\} $$
View solution Problem 200
$$ \left.\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \sin 5 x}{x^{2} \sin 3 x} \text { . \\{Ans. } \frac{10}{3}\right\\} $$
View solution Problem 202
$$ \lim _{x \rightarrow x}\left(\frac{x-1}{x+1}\right)^{x} \cdot\left\\{\text { Ans. } e^{-2}\right\\} $$
View solution Problem 203
$$ \lim _{x \rightarrow \infty}\left(\frac{x^{2}+6}{x^{2}-6}\right)^{x} \cdot\\{\text { Ans. } 1\\} $$
View solution