lim_(x rarr3)(x-3)/(2-sqrt(x+1))=

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Substitute x with 3: Substitute the value of x with 3 in the expression to check if the limit can be directly calculated. lim_(x -> 3)(x-3)/(2-sqrt(x+1)) = (3-3)/(2-sqrt(3+1))Simplify after substitution: Simplify the expression after substitution to see if it results in an indeterminate form. (3-3)/(2-sqrt(3+1)) = 0/(2-sqrt(4)) = 0/0 Since we get 0/0, this is an indeterminate form, and we cannot directly calculate the limit.Apply L'Hôpital's Rule: Apply L'Hôpital's Rule because we have an indeterminate form of 0/0. This rule states that if the limit as x approaches a of f(x)/g(x) is 0/0 or ∞/∞, then the limit is the same as the limit of the derivatives of the numerator and the denominator, provided that the limit of the derivatives exists. lim_(x -> 3)(x-3)/(2-sqrt(x+1)) = lim_(x -> 3)(d/dx(x-3))/(d/dx(2-sqrt(x+1)))Differentiate numerator and denominator: Differentiate the numerator and the denominator with respect to x. The derivative of the numerator (x-3) with respect to x is 1. The derivative of the denominator (2-sqrt(x+1)) with respect to x is -1/(2sqrt(x+1)). So, lim_(x -> 3)(d/dx(x-3))/(d/dx(2-sqrt(x+1))) = lim_(x -> 3)(1)/(-1/(2sqrt(x+1)))Simplify after differentiation: Simplify the expression after differentiation. lim_(x -> 3)(1)/(-1/(2sqrt(x+1))) = lim_(x -> 3)(-2sqrt(x+1))Substitute x with 3: Substitute x with 3 in the simplified expression after applying L'Hôpital's Rule. lim_(x -> 3)(-2sqrt(x+1)) = -2sqrt(3+1) = -2sqrt(4) = -2*2 = -4

$\lim _{x \rightarrow 3} \frac{x-3}{2-\sqrt{x+1}}=$

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lim⁡x→3x−32−x+1= \lim _{x \rightarrow 3} \frac{x-3}{2-\sqrt{x+1}}= limx→3​2−x+1​x−3​=

$\lim _{x \rightarrow 3} \frac{x-3}{2-\sqrt{x+1}}=$