IA:MuliaAilum

lim

(x²-x+ 2)(2x + 3)

x → 2 √x² +5-√√3x +3

To solve this limit, we need to simplify the expressions inside the parentheses and simplify the radical expressions.

First, let's simplify the expressions inside the parentheses of the first term:

x² - x + 2 = (x - 2)(x - 1) + 4

Next, let's simplify the radical expressions in the second term:

√x² + 5 - √√3x + 3 = √(x² + 5) - √(√3x + 3)

To simplify further, we can use the conjugate of the second radical expression:

√(√3x + 3) = √3(x + 1) = √3 * √(x + 1)

So, the conjugate is: √(√3x + 3) + 1 = √3 * √(x + 1) + 1

Multiplying the numerator and denominator by the conjugate of the second term, we get:

lim [(x - 2)(x - 1) + 4)(2x + 3)] / [(√(x² + 5) - √3 * √(x + 1)) * (√(√3x + 3) + 1)]

x → 2

Plugging in x = 2, we get:

[(2 - 2)(2 - 1) + 4)(2(2) + 3)] / [(√(2² + 5) - √3 * √(2 + 1)) * (√(√3(2) + 3) + 1)]

= (4)(7) / [(√9 - √3 * √3) * (√(√6 + 1))]

= 28 / (2 * √(√6 + 1))

= 14 / √(√6 + 1)

Therefore, the limit is 14 / √(√6 + 1).