solve this---If $\lim_{x \to 0} \frac{x(1 - a \cos x) + b \sin x}{x^3} = \frac{1}{3}$,
find $a$ and $b$.
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We have a limit problem where we need to find the values of parameters a and b. The given limit equals one-third as x approaches zero. This creates a zero over zero indeterminate form, which we can solve using L'Hôpital's rule or Taylor series expansion. We'll use the Taylor series approach to systematically find the parameter values.
We'll use Taylor series expansion to solve this limit. First, we expand cosine x as 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial, and so on. Similarly, sine x expands as x minus x cubed over 3 factorial plus higher order terms. We substitute these expansions into our numerator expression x times 1 minus a cosine x plus b sine x. This transforms our expression into a polynomial in x divided by x cubed.
After expanding the Taylor series, we get x times a x squared over 2 plus b x minus b x cubed over 6, plus higher order terms. Collecting terms by powers of x, we have b x plus a x cubed over 2 minus b x cubed over 6. For the limit to exist and equal one-third, the coefficients of x and x squared must be zero, since we're dividing by x cubed.
From our coefficient analysis, we need the coefficient of x to be zero, which gives us b equals zero. However, for the limit to equal one-third, we need the remaining x cubed term divided by x cubed to equal one-third. This means negative b over 6 equals one-third, giving us b equals negative 2. But this contradicts our earlier finding that b equals zero. We need to reconsider our approach.
We need to find the values of a and b such that the limit of x times one minus a cosine x plus b sine x divided by x cubed equals one third as x approaches zero.
To solve this limit, we need the Taylor series expansions. Cosine x equals 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial and so on. Sine x equals x minus x cubed over 3 factorial plus x to the fifth over 5 factorial and so on.
Let's expand the numerator step by step. First, 1 minus a cosine x becomes 1 minus a plus a x squared over 2. When multiplied by x, we get 1 minus a times x plus a x cubed over 2. For b sine x, we get b x minus b x cubed over 6.
Combining all terms, we get 1 minus a plus b times x plus a over 2 minus b over 6 times x cubed. When divided by x cubed, this becomes 1 minus a plus b over x squared plus a over 2 minus b over 6.
For the limit to exist and equal one third, the coefficient of 1 over x squared must be zero, so 1 minus a plus b equals zero. Also, a over 2 minus b over 6 must equal one third. From the first equation, b equals a minus 1. Substituting into the second equation and solving, we get a equals one half and b equals negative one half.