Problem 313 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \lim _{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} .\left\\{\text { Ans. } \frac{1}{8 \sqrt{3}}\right\\} $$

Step-by-Step Solution

Answer

The short answer based on the step-by-step solution is: \(\lim_{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} = \frac{1}{8\sqrt{3}}\)

Step 1: 1. Identify the function and rewrite the limit notation

We are given the function: \(\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\) We want to find the limit as x approaches 2: \(\lim_{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\)

Step 2: 2. Rationalize the numerator

To rationalize the numerator, we will multiply both the numerator and denominator by the conjugate of the numerator: \(\lim _{x \rightarrow 2} \frac{\left(\sqrt{1+\sqrt{2+x}}-\sqrt{3}\right)\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}{(x-2)\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}\)

Step 3: 3. Simplify the expression

Multiply the numerator using the difference of squares: \(\lim _{x \rightarrow 2} \frac{(1+\sqrt{2+x})-3}{(x-2)\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}\) Now, simplify the expression in the numerator: \(\lim_{x \rightarrow 2} \frac{\sqrt{2+x} -2}{(x-2)\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}\)

Step 4: 4. Divide both the numerator and denominator by (x-2)

This step will help us cancel out the (x-2) term in the denominator, as follows: \(\lim_{x \rightarrow 2} \frac{\frac{\sqrt{2+x}-2}{x-2}}{\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}\)

Step 5: 5. Use the limit definition for derivatives

Now we can rewrite the limit in the numerator as a derivative of a function: \(\lim_{x \rightarrow 2} \frac{\frac{d}{dx}\left(\sqrt{2+x}\right)}{\left(\sqrt{1+\sqrt{2+x}}+\sqrt{3}\right)}\) The derivative of \(\sqrt{2+x}\) with respect to x is: \(\frac{1}{2\sqrt{2+x}}\)

Step 6: 6. Plug in x = 2 and find the limit

Now we can substitute x = 2 in the limit: \(\lim_{x \rightarrow 2} \frac{\frac{1}{2\sqrt{2+2}}}{\left(\sqrt{1+\sqrt{2+2}}+\sqrt{3}\right)} = \frac{\frac{1}{4}}{\left(\sqrt{1+\sqrt{4}}+\sqrt{3}\right)}\) This simplifies to: \(\frac{1}{4\left(\sqrt{1+2}+\sqrt{3}\right)}=\frac{1}{8\sqrt{3}}\) So, the limit of the given function as x approaches 2 is: \(\lim_{x \rightarrow 2} \frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} = \frac{1}{8\sqrt{3}}\)

Key Concepts

Rationalizing Techniques

Rationalizing is a technique used to simplify expressions, particularly in limits and fractions where roots are involved. The main goal of rationalizing is to eliminate these roots, usually focusing on the numerator or denominator.
To achieve this, we often multiply the expression by the 'conjugate'. The conjugate of a binomial with a square root, such as \(a - b\), is \(a + b\). By multiplying by this conjugate, we exploit the difference of squares formula, which helps remove the square roots. This process can simplify the expression significantly and make it easier to evaluate limits especially when direct substitution leads to an indeterminate form like \(\frac{0}{0}\).

• Multiply both top and bottom by the conjugate of the term with the square root.
• Apply the difference of squares formula to simplify further: \((a-b)(a+b) = a^2 - b^2\).
• This allows the new denominator or numerator to be resolved, revealing components that cancel with other terms.

In the original exercise, multiplying by the conjugate of \(\sqrt{1+\sqrt{2+x}}-\sqrt{3}\) allowed for a straightforward simplification of the numerator. Thus, the limit could be found much more easily.

Derivative Definition

The derivative is a fundamental concept in calculus. It helps us understand rates of change. Specifically, it describes how a function changes as its input changes. The derivative of a function \(f(x)\) at a point \(x\) is denoted as \(f'(x)\) and can be defined using limits.
The formal definition of a derivative is:

• \( f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \)

This represents the instantaneous rate of change of the function, which is geometrically interpreted as the slope of the tangent line to the curve at point \(x\).
In the original exercise, the limit process essentially rewrote part of the expression as a derivative of the function \(\sqrt{2+x}\). For \(\sqrt{2+x}\), the derivative is \(\frac{1}{2\sqrt{2+x}}\), showcasing that the limit processes can intertwine deeply with derivatives.

Difference of Squares

The difference of squares is a powerful algebraic tool that simplifies expressions involving subtracted squares. This formula is expressed as:
\(a^2 - b^2 = (a - b)(a + b)\)
It helps by turning two squared terms separated by a subtraction into a product. This is incredibly useful in calculus, particularly concerning rationalizing, as it simplifies expressions involving square roots.
Here’s why it works effectively:

• The expression \(a^2 - b^2\) factors into two binomials \((a-b)(a+b)\).
• When applied, square terms dissolve leaving a much simpler algebraic form, often removing impediments like indeterminate expressions.

In solving the limit from the original exercise, the application of the difference of squares helped eliminate the square roots in the numerator following the rationalization step. Simplifying the expression further allowed for easy cancellation of the (x-2) factor in the denominator, thus making it possible to evaluate the limit.