If limit as delta x approaches 0 of tan(0+Δx)-tan(0)/Δx =1
which of the following is false:
d/dx [tanx]=1
the slope of y = tan(x) at x = 0 is 1
y = tan(x) is continuous at x = 0
y = tan(x) is differentiable at x = 0
assuming those 0's should be x's
the first assertion is false. d/dx(tanx) = 1 at x=0, not everywhere.
Kind of an odd problem, since the last 3 statements are all obviously true.
To determine which statement is false, let's analyze each option:
1. d/dx [tanx] = 1:
To find the derivative of tan(x), we can use standard rules of differentiation. The derivative of tan(x) is sec^2(x). Since sec^2(0) = 1, this statement is true.
2. The slope of y = tan(x) at x = 0 is 1:
To find the slope of the tangent line to y = tan(x) at x = 0, we need to take the derivative of tan(x) and evaluate it at x = 0. As mentioned earlier, the derivative of tan(x) is sec^2(x), and sec^2(0) = 1. Therefore, the slope at x = 0 is 1. This statement is also true.
3. y = tan(x) is continuous at x = 0:
To check if y = tan(x) is continuous at x = 0, we need to evaluate the limit from the left and the right and see if they match. In this case, the limit as x approaches 0 from the left is tan(-π/2) which is negative infinity. The limit as x approaches 0 from the right is tan(π/2) which is positive infinity. Since the left and right limits are not equal, y = tan(x) is not continuous at x = 0. Thus, this statement is false.
4. y = tan(x) is differentiable at x = 0:
To determine if y = tan(x) is differentiable at x = 0, we need to check if the derivative exists at that point. As we mentioned earlier, the derivative of tan(x) is sec^2(x), which is defined for all real values of x, including x = 0. Therefore, y = tan(x) is differentiable at x = 0. This statement is true.
Based on the analysis, statement 3, "y = tan(x) is continuous at x = 0," is false.