how do you find the limit of the square root of x plus 3 minus 3 all divided by x minus 6 when x is approaching 6
+128406 lim as x → 6 of [√(x +3 ) - 3] / [ x - 6]
Note that this gives us the indeterminate form... ( 0 / 0 ).... when we take the limit !!!
When we have this situation, we can try to employ something called L'Hopital's Rule to find the limit.
Take the derivative of the numerator and the denominator.....this gives us
[(1/2)*(x + 3)^(-1/2)] / [1] = (1/2)*(x + 3)^(-1/2)
Now, take the limit of this as x → 6 so we have
[(1/2)*(6 + 3)^(-1/2) = (1/2)*(9)^(-1/2) = (1/2)*(1/3) = 1/6
Note that the function doesn't actually exist at x = 6.......but, we don't care about that......we only care about what happens as we approach 6 from both sides.......!!!!!
+128406 lim as x → 6 of [√(x +3 ) - 3] / [ x - 6]
Note that this gives us the indeterminate form... ( 0 / 0 ).... when we take the limit !!!
When we have this situation, we can try to employ something called L'Hopital's Rule to find the limit.
Take the derivative of the numerator and the denominator.....this gives us
[(1/2)*(x + 3)^(-1/2)] / [1] = (1/2)*(x + 3)^(-1/2)
Now, take the limit of this as x → 6 so we have
[(1/2)*(6 + 3)^(-1/2) = (1/2)*(9)^(-1/2) = (1/2)*(1/3) = 1/6
Note that the function doesn't actually exist at x = 6.......but, we don't care about that......we only care about what happens as we approach 6 from both sides.......!!!!!
