Problem 276 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow \pm \infty}\left(\frac{x+1}{2 x-1}\right)^{x} \text { \\{ Ans. } 0,+\infty\right\\} $$

Step-by-Step Solution

Answer

Hence, the answer of the limit \( \lim _{x \rightarrow +\infty}\left(\frac{x+1}{2x-1}\right)^{x} \) is 0, and \( \lim _{x \rightarrow -\infty}\left(\frac{x+1}{2x-1}\right)^{x} \) is \( +\infty \).

Step 1: Rewrite the limit

Rewrite the expression \( \left(\frac{x+1}{2x-1}\right)^{x} \) in the form of \( e^{x \cdot ln\left(\frac{x+1}{2x-1} \right)} \) to have a different perspective of the same limit.

Step 2: Handle infinity times zero form

The limit \( \lim _{x \rightarrow \pm \infty}(x \cdot ln\left(\frac{x+1}{2x-1} \right)) \) now has the indeterminate form \( \infty \cdot 0 \). This can be overcome by rewriting the expression into fractions. Consider \( \lim _{x \rightarrow \pm \infty}\left( \frac{ln \left(\frac{x+1}{2x-1} \right)}{1/x} \right) \). This appears as \( \frac{0}{0} \) when \( x \rightarrow \pm \infty \).

Step 3: Apply L'Hopital's Rule

With the form representing \( 0/0 \), L'Hopital's rule can be applied which is \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \). Accordingly, take derivative of the numerator and the denominator of the fraction separately yielding \( \frac{\frac{2x-1-(x+1)}{(2x-1)²}}{-1/x²} \). After simplification, the expression becomes \( \frac{1}{(2x-1)²} \).

Step 4: Compute the limit

Now compute the limit \( \lim _{x \rightarrow \pm \infty}\frac{1}{(2x-1)²} \). For \( x \rightarrow +\infty \), the limit is 0. For \( x \rightarrow -\infty \), the limit is \( +\infty \) as the denominator turns positive due to the square.

Key Concepts

Indeterminate Forms

In calculus, an indeterminate form arises when substituting values into a function leads to undefined expressions. These expressions are often encountered when determining limits. Common types of indeterminate forms include:

• \( \frac{0}{0} \)
• Infinitive differences such as \( \infty - \infty \)
• Infinite multiplication like \( \infty \cdot 0 \)

Meet one of the most interesting cases in the example exercise. The limit \(\lim _{x \rightarrow \pm \infty}(x \cdot \ln\left(\frac{x+1}{2x-1} \right))\)leads to an indeterminate form of \( \infty \cdot 0 \). This form invites the use of creative algebraic manipulation to set the stage for limit resolution. Transforming this expression into a fraction form, it becomes \(\frac{\ln(\frac{x+1}{2x-1})}{1/x}\),which then can be evaluated using L'Hopital's Rule.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus for solving limit problems involving indeterminate forms such as \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). The rule states that if the limits of \( f(x) \) and \( g(x) \) both approach 0 or both approach infinity as \( x \) approaches a point or infinity, then:\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]It's important to note that L'Hopital’s Rule can only be applied when the functions are differentiable near the point of interest. For instance, in our original problem, after rewriting into a fraction:\(\frac{\ln(\frac{x+1}{2x-1})}{1/x}\)forms a 0/0 indeterminate state. By differentiating the numerator and the denominator individually, we simplify the expression and can directly find the limit as \( x \) approaches \( \pm \infty \). This often makes handling otherwise complex limits far simpler.

Infinity in Calculus

Infinity, denoted as \( \infty \), plays a crucial role in calculus. It represents an idea rather than a number: something without bound or limit. In terms of limit problems, infinity helps describe behavior as a variable grows very large or very small.In calculus:

• \( x \rightarrow +\infty \) indicates that \( x \) increases without bound.
• \( x \rightarrow -\infty \) denotes that \( x \) decreases without bound.

For the exercise under analysis, solving\(\lim_{x \rightarrow \pm \infty} \frac{x+1}{2x-1}\)involves examining the numerator and the denominator's behavior as \( x \) gets extremely large or small. This requires understanding how these expressions relate to the concept of infinity. Ultimately, this concept aids in recognizing the limit as either \( 0 \) or \( +\infty \), depending on the direction of \( x \).