uu) lim ln(1-x) as x->1-
yy) lim (√(6-x)-2)/(√(3-x)-1) as x-> 2
zz) lim (1-(1/2)arctanx) as x->-∞
bbb) lim ln|1-x| as x->1
If these are hard to interpret with all the parentheses, you can plug them in to Wolfram Alpha and it interprets them perfectly.
Sorry, I forgot one:
w) lim (1-2cos^2x)/(cosx-sinx) as x->π/4
if you've already been to wolfram, and it shows steps as needed, why bother posting here?
To find the limits in these expressions, we can use some limit properties and techniques. Let's go through each question step by step:
uu) To find the limit of ln(1-x) as x approaches 1 from the left (1-), we can use the fact that the natural logarithm function is continuous. Therefore, we can directly substitute x = 1 into the function. The limit is thus:
lim(x->1-) ln(1-x) = ln(1-1) = ln(0) = -∞
yy) To find the limit of (√(6-x)-2)/(√(3-x)-1) as x approaches 2, we can try to directly substitute x = 2 into the expression. However, this expression results in an indeterminate form of 0/0.
To resolve this, we can simplify the expression using the conjugate. Multiply the numerator and denominator by the conjugate of the denominator (√(3-x)+1):
lim(x->2) (√(6-x)-2)/(√(3-x)-1) * (√(3-x)+1)/(√(3-x)+1) = lim(x->2) (6-x-4)/(3-x-1) = lim(x->2) (2-x)/(2-x) = lim(x->2) 1 = 1
zz) To find the limit of 1 - (1/2)arctan(x) as x approaches negative infinity, we can consider the behavior of the arctan(x) function as x becomes negative infinity. The arctan(x) function approaches -π/2.
So, the limit becomes:
lim(x->-∞) 1 - (1/2)arctan(x) = 1 - (1/2)(-π/2) = 1 + π/4
bbb) To find the limit of ln|1-x| as x approaches 1, we can consider the behavior of the absolute value function |1-x| as x approaches 1. The absolute value function simplifies to 1 - x for x < 1 and x - 1 for x > 1.
Since we are taking the limit as x approaches 1, we consider x from the left (1-) and the right (1+) separately:
lim(x->1-) ln|1-x| = lim(x->1-) ln(1 - x) = ln(1 - 1) = ln(0) = -∞
lim(x->1+) ln|1-x| = lim(x->1+) ln(x - 1) = ln(1 - 1) = ln(0) = -∞
Therefore, the limit of ln|1-x| as x approaches 1 does not exist.
While Wolfram Alpha can be a useful tool to check the results and provide quick interpretations, it is also helpful to understand the process behind finding these limits.