Answer:
The idea is to transform the expression by multiplying 
with its conjugate, 
.
Step-by-step explanation:
For any real number 
and 
, 
.
The factor is irrational. However, when multiplied with its square root conjugate , the product would become rational:
.
The idea is to multiply 
by 
so as to make it easier to take the limit.
Since 
, multiplying the expression by this fraction would not change the value of the original expression.
![\begin{aligned} & \lim\limits_{x \to \infty} \sqrt{x} \, (\sqrt{x + 1} - \sqrt{x}) \\ &= \lim\limits_{x \to \infty} \left[\sqrt{x} \, (\sqrt{x + 1} - \sqrt{x})\cdot \frac{\sqrt{x + 1} + \sqrt{x}}{\sqrt{x + 1} + \sqrt{x}}\right] \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}\, ((x + 1) - x)}{\sqrt{x + 1} + \sqrt{x}} \\ &= \lim\limits_{x \to \infty} \frac{\sqrt{x}}{\sqrt{x + 1}+ \sqrt{x}}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%26%20%5Clim%5Climits_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Csqrt%7Bx%7D%20%5C%2C%20%28%5Csqrt%7Bx%20%2B%201%7D%20-%20%5Csqrt%7Bx%7D%29%20%5C%5C%20%26%3D%20%5Clim%5Climits_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cleft%5B%5Csqrt%7Bx%7D%20%5C%2C%20%28%5Csqrt%7Bx%20%2B%201%7D%20-%20%5Csqrt%7Bx%7D%29%5Ccdot%20%5Cfrac%7B%5Csqrt%7Bx%20%2B%201%7D%20%2B%20%5Csqrt%7Bx%7D%7D%7B%5Csqrt%7Bx%20%2B%201%7D%20%2B%20%5Csqrt%7Bx%7D%7D%5Cright%5D%20%5C%5C%20%26%3D%20%5Clim%5Climits_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%7Bx%7D%5C%2C%20%28%28x%20%2B%201%29%20-%20x%29%7D%7B%5Csqrt%7Bx%20%2B%201%7D%20%2B%20%5Csqrt%7Bx%7D%7D%20%5C%5C%20%26%3D%20%5Clim%5Climits_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%7Bx%7D%7D%7B%5Csqrt%7Bx%20%2B%201%7D%2B%20%5Csqrt%7Bx%7D%7D%5Cend%7Baligned%7D)
.
The order of 
in both the numerator and the denominator are now both 
. Hence, dividing both the numerator and the denominator by 
(same as 
) would ensure that all but the constant terms would approach 
under this limit:
.
By continuity:
.
Answer:
0.21
Step-by-step explanation:
Answer:
Step-by-step explanation:
Because MK is a diameter, then angle L is a right angle. We already know that the measure of angle K is 50, so the measure of angle M has to be 40 because of the triangle angle-sum theorem. The rule for inscribed angles and the arcs they cut off is that the angle is half the measure of its intercepted arc or, likewise, the arc is twice the measure of the angle that cuts it off. Since arc LK is across from angle M and is cut off by angle M, then arc LK is twice the measure of angle M, and is 80. That's the same reason why angle L is 90; arc MK is a semi-circle, with a degree measure of 180, and angle L is half of that.
Arc LK = 80
