Evaluate the limit of the numerator.

To solve this limit, we will evaluate the numerator and the denominator separately as x . Step 1: Evaluate the limit of the numerator. Let N(x) = sqrt(x^2+2x+3) - sqrt(x^2+3). This is an indeterminate form of type - . We will multiply by the conjugate expression: N(x) = (sqrt(x^2+2x+3) - sqrt(x^2+3))(sqrt(x^2+2x+3) + sqrt(x^2+3))sqrt(x^2+2x+3) + sqrt(x^2+3) N(x) = ((x^2+2x+3) - (x^2+3))/(sqrt(x^2+2x+3) + x^2+3) N(x) = (2x)/(sqrt(x^2+2x+3) + x^2+3) To evaluate this limit as x , we divide the numerator and the denominator by x (noting that sqrt(x^2) = x for x > 0): N(x) = (2)/(sqrt(x^2+2x+3)x^2) + sqrt((x^2+3)/(x^2)) N(x) = (2)/(sqrt(1+2)x)+(3)/(x^2) + sqrt(1+(3)/(x^2)) Now, we can evaluate the limit as x : _x N(x) = (2)/(sqrt(1+0+0) + 1+0) = (2)/(1+1) = 1 Step 2: Evaluate the limit of the denominator. Let D(x) = 1 - (^-1(((1)/(x)))). Let u = (1)/(x). As x , u 0^+. The denominator becomes: D(u) = 1 - (^-1((u))) As u 0^+: • (u) (0) = 0. • ^-1((u)) ^-1(0) = 0. • (^-1((u))) (0) = 1. So, _u 0^+ D(u) = 1 - 1 = 0. Step 3: Determine the sign of the denominator as it approaches 0. We use equivalent functions for u 0^+: • (u) u • ^-1(v) v (for v 0) So, ^-1((u)) (u) u. Let w = ^-1((u)). Then w u as u 0^+. • 1 - (w) (w^2)/(2) (for w 0) Substituting w u, we get: D(u) (u^2)/(2). Since u = (1)/(x), we have D(x) ((1/x)^2)/(2) = (1)/(2x^2). As x , (1)/(2x^2) is always positive. Therefore, the denominator approaches 0 from the positive side (0^+). Step 4: Combine the limits of the numerator and the denominator. The original limit is _x (N(x))/(D(x)). We found _x N(x) = 1. We found _x D(x) = 0^+. Therefore, the limit is of the form (1)/(0^+), which tends to positive infinity. _x sqrt(x^2+2x+3) - sqrt(x^2+3)1 - (^-1(((1)/(x)))) = (1)/(0^+) = The final answer is . Got more? Send 'em!