If Lim x-->3 (√2x+3 -x/(√x+1)-x+1)^(x-1-√x^2-5/x^2-5x+6) can be expressed in the form of a√b/c where a,b,c belongs to Me , then find least value of a^2+b^2+c^2

To solve the limit as \( x \) approaches 3 for the expression \( \left( \frac{\sqrt{2x+3} - x}{\sqrt{x+1}} - x + 1 \right)^{\left( x - 1 - \frac{\sqrt{x^2 - 5}}{x^2 - 5x + 6} \right)} \), we need to analyze both the base and the exponent separately. Let's break this down step by step.

Step 1: Analyzing the Base

First, we look at the base of the expression:

\[ \frac{\sqrt{2x+3} - x}{\sqrt{x+1}} - x + 1 \]

As \( x \) approaches 3, we can substitute \( x = 3 \) directly into the base:

\[ \sqrt{2(3)+3} - 3 = \sqrt{9} - 3 = 3 - 3 = 0 \]

Now, substituting \( x = 3 \) into \( \sqrt{x+1} \):

\[ \sqrt{3+1} = \sqrt{4} = 2 \]

Thus, the base becomes:

\[ \frac{0}{2} - 3 + 1 = 0 - 3 + 1 = -2 \]

So, the base approaches -2 as \( x \) approaches 3.

Step 2: Analyzing the Exponent

Next, we need to evaluate the exponent:

\[ x - 1 - \frac{\sqrt{x^2 - 5}}{x^2 - 5x + 6} \]

First, substituting \( x = 3 \):

\[ 3 - 1 - \frac{\sqrt{3^2 - 5}}{3^2 - 5(3) + 6} = 2 - \frac{\sqrt{9 - 5}}{9 - 15 + 6} = 2 - \frac{\sqrt{4}}{0} \]

Here, we encounter a division by zero, indicating that we need to analyze this limit more carefully. We can use L'Hôpital's Rule or algebraic manipulation to resolve this.

Step 3: Simplifying the Exponent

To simplify the expression, we can rewrite the denominator:

\[ x^2 - 5x + 6 = (x-2)(x-3) \]

As \( x \) approaches 3, the denominator approaches 0, and we need to find the limit of the numerator:

\[ \sqrt{x^2 - 5} = \sqrt{(x-3)(x+3)} \]

Using L'Hôpital's Rule, we differentiate the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx}(\sqrt{x^2 - 5}) = \frac{x}{\sqrt{x^2 - 5}}, \quad \text{Denominator: } \frac{d}{dx}((x-2)(x-3)) = 2x - 5 \]

Evaluating the limit gives us:

\[ \lim_{x \to 3} \frac{\sqrt{x^2 - 5}}{(x-2)(x-3)} = \lim_{x \to 3} \frac{x}{\sqrt{x^2 - 5}(2x - 5)} = \frac{3}{\sqrt{4}(1)} = \frac{3}{2} \]

Step 4: Final Limit Calculation

Now we can substitute this back into our exponent:

\[ 2 - \frac{3}{2} = \frac{4}{2} - \frac{3}{2} = \frac{1}{2} \]

Thus, the limit can be expressed as:

\[ (-2)^{\frac{1}{2}} = \sqrt{-2} = i\sqrt{2} \]

Expressing in the Required Form

We need to express \( i\sqrt{2} \) in the form \( \frac{a\sqrt{b}}{c} \). Here, we can take \( a = 1 \), \( b = 2 \), and \( c = 1 \). Now, we calculate:

\[ a^2 + b^2 + c^2 = 1^2 + 2^2 + 1^2 = 1 + 4 + 1 = 6 \]

Therefore, the least value of \( a^2 + b^2 + c^2 \) is 6.