Identify Problem: We are asked to find the limit of the function (1−cosx)/(1+x2−1−x2) as x approaches 0. This is a limit problem involving trigonometric and square root functions.
Recognize Indeterminate Form: First, we notice that direct substitution of x=0 into the function gives us a 0/0 indeterminate form. This means we need to apply some algebraic manipulation or a limit law to evaluate the limit.
Apply Algebraic Manipulation: To resolve the indeterminate form, we can multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of 1+x2−1−x2 is 1+x2+1−x2.
Multiply by Conjugate: Multiplying the numerator and denominator by the conjugate, we get: limx→0(1+x2−1−x2)(1+x2+1−x2)(1−cosx)(1+x2+1−x2)
Simplify Denominator: Simplify the denominator using the difference of squares formula: (a−b)(a+b)=a2−b2. This gives us:x→0lim(1+x2)−(1−x2)(1−cosx)(1+x2+1−x2)
Combine Like Terms: Simplify the denominator further: limx→01+x2−1+x2(1−cosx)(1+x2+1−x2)
Distribute and Simplify Numerator: Combine like terms in the denominator: limx→02x2(1−cosx)(1+x2+1−x2)
Evaluate Each Term: Now, we can simplify the numerator by distributing (1−cosx) over the sum inside the square roots: x→0lim2x2(1−cosx)1+x2+(1−cosx)1−x2
Correct Error and Re-evaluate: We can now evaluate the limit of each term separately as x approaches 0. For the first term, (1−cosx)1+x2, as x approaches 0, (1−cosx) approaches 0 and 1+x2 approaches 1. For the second term, (1−cosx)1−x2, as x approaches 0, (1−cosx) approaches 0 and 04 approaches 1.
Correct Error and Re-evaluate: We can now evaluate the limit of each term separately as x approaches 0. For the first term, (1−cosx)1+x2, as x approaches 0, (1−cosx) approaches 0 and 1+x2 approaches 1. For the second term, (1−cosx)1−x2, as x approaches 0, (1−cosx) approaches 0 and 04 approaches 1.However, we realize that there is a mistake in the previous step. The limit of (1−cosx) as x approaches 0 is not 0, but rather (1−cosx)1+x20. This means that the numerator approaches 0 as x approaches 0. We need to correct this error and re-evaluate the limit.
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