Problem 110 - A Complete Resource Book in Mathematics for JEE Main

Question

We know that if \(\lim _{x \rightarrow a} f(x)=l\) and \(\lim _{x \rightarrow a} g(x)=m(\neq 0)\), then $$ \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _{x \rightarrow a} f(x)}{\lim _{x \rightarrow a} g(x)} $$ However, if \(\lim _{x \rightarrow a} g(x)=0=\lim _{x \rightarrow a} f(x)\), we cannot say anything definite about the existence of \(\lim _{x \rightarrow a} \frac{f(x)}{g(x)}\). Though in some cases this limit exists. Any expression of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) is termed as an indeterminate form. Many other expressions like \(\infty-\infty, 1^{\infty}, \infty^{0}, 0^{0}, 0 \times \infty\) which can be reduced to the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) are also called indeterminate forms. If \(\frac{f(x)}{g(x)}\) is indeterminate at \(x=a\) of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then $$ \lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)} $$ where \(f^{\prime}\) is derivative of \(f\). If \(\frac{f^{\prime}(x)}{g^{\prime}(x)}\), too, is indeterminate at \(x=a\) of the type \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim _{x \rightarrow a} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}\) This can be continued till we finally arrive at a determinate result. If \(\lim _{x \rightarrow 0} \frac{\sin 2 x+a \sin x}{x^{3}}\) be finite, then the value of \(a\) and the limit are given by (A) \(-2,1\) (B) \(-2,-1\) (C) 2,1 (D) \(2,-1\)

Step-by-Step Solution

Answer

The value of \(a\) is \(-2\), and the limit is \(0\).

Step 1: Understand the Given Problem

We need to evaluate \( \lim_{x \rightarrow 0} \frac{ \sin 2x + a \sin x}{x^3} \) given that the limit is finite and determine the value of \(a\) and the limit. Since both the numerator and denominator approach zero as \(x\rightarrow 0\), it is an indeterminate form \(\frac{0}{0}\).

Step 2: Apply L'Hopital's Rule

Since \(\frac{0}{0}\) is an indeterminate form, we can apply L'Hopital's rule which states that \[\lim_{x \rightarrow 0} \frac{ \sin 2x + a \sin x}{x^3} = \lim_{x \rightarrow 0} \frac{ (\sin 2x + a \sin x)'}{(x^3)'} \]First, we calculate the derivatives: - Derivative of the numerator \((\sin 2x + a \sin x)' = 2\cos 2x + a\cos x \)- Derivative of the denominator \((x^3)' = 3x^2 \)

Step 3: Get New Limit After First Derivative

Applying L'Hopital's rule once gives us the new limit:\[\lim_{x \rightarrow 0} \frac{ 2\cos 2x + a\cos x }{3x^2}\]This expression is still \(\frac{0}{0}\) because \(\cos(0) = 1\), so we apply L'Hopital's rule again.

Step 4: Apply L'Hopital's Rule Again

Take second derivatives of the numerator and the denominator:- Second derivative of the numerator: \((2\cos 2x + a\cos x)' = -4\sin 2x - a\sin x\)- Second derivative of the denominator: \((3x^2)' = 6x\)Then the limit becomes:\[\lim_{x \rightarrow 0} \frac{-4\sin 2x - a\sin x}{6x} \]This is still \(\frac{0}{0}\), so we need one more derivative.

Step 5: Third Application of L'Hopital's Rule

Now calculate the third derivatives:- Third derivative of the numerator: \((-4\sin 2x - a\sin x)' = -8\cos 2x - a\cos x\)- Third derivative of the denominator: \((6x)' = 6\)Therefore:\[\lim_{x \rightarrow 0} \frac{-8\cos 2x - a\cos x}{6}\]Evaluate this limit: at \(x=0\), \(\cos 0 = 1\), so we plug \(x=0\) directly into the expression.

Step 6: Determine Value of a and the Limit

We have:\[\lim_{x \rightarrow 0} \frac{-8(1) - a(1)}{6} = \frac{-8 - a}{6} \]For this limit to be finite, it must equal zero:\[\frac{-8 - a}{6} = 0 \]Solving \(-8 - a = 0\) gives \(a = -8\). Substituting \(a = -2\), not zero, the remainder \(-2\) of L'Hopital gives zero, as indeterminate does not continue beyond.

Key Concepts

Indeterminate Forms

Sometimes in calculus, we encounter limits that result in forms where we cannot readily determine their value. These are known as "Indeterminate Forms". A classic example is \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), both of which do not provide obvious numerical results. Other indeterminate forms include combinations like \(\infty-\infty\), \(1^{\infty}\), \(\infty^{0}\), \(0^{0}\), and \(0 \times \infty\). Recognizing these forms is crucial for applying the right calculus tools for evaluation.

Indeterminate forms suggest that simply substituting the limiting value into the function's terms doesn't work. Instead, you'll often need to manipulate the function algebraically or apply special calculus methods like L'Hopital's Rule to resolve the limit. Indeterminate forms require careful handling to determine the precise limiting value.

Calculus Limits

The concept of limits is fundamental in calculus. Limits allow us to determine the behavior of functions as their inputs approach certain values. It captures the idea of approaching a specified point, usually expressed as \( \lim_{x \rightarrow a} f(x)\).

These limits give us insights into the continuity and behavior of functions near important points. Some limits are immediate, such as when \(\lim_{x \rightarrow a} f(x) = l\) directly produces a number. However, things become interesting when we encounter limits producing indeterminate forms. In such cases, we need additional steps or tools to find the precise limit value.

• Limits are used to define key calculus concepts like derivatives and integrals.
• Understanding limits is essential for grasping L'Hopital’s Rule, which helps resolve difficult indeterminate limits.

Overall, mastering calculus limits is crucial for understanding advanced calculus topics and applications.

Derivatives

Derivatives represent the rate of change of a function with respect to a variable. In simple terms, it's like determining the slope of a curve at any given point. Derivatives are denoted as \( f'(x) \) or \( \frac{df}{dx} \).

Derivatives play a key role in L'Hopital's Rule, which we use to evaluate limits that initially result in indeterminate forms. In such cases, we take the derivative of the numerator and denominator separately, then reevaluate the limit. If the expression is still indeterminate, further derivatives are taken until a determinate form is reached.

Key points about derivatives include:

• They help in understanding how functions behave, change, and respond to inputs.
• Derivatives are used in various applications like physics for acceleration and velocity calculations.
• Knowing how to compute derivatives is essential for applying L'Hopital's Rule effectively.

In short, derivatives are about understanding how a function changes and are crucial tools for solving calculus problems.