The Maclaurin expansion of \( \sin x \) begins
Answer: D
The derivatives at the origin cycle through \( 1, 0, -1, 0 \), so only odd powers survive and the signs alternate:
\[ \sin x \;=\; x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \cdots \;=\; x - \frac{x^{3}}{6} + \frac{x^{5}}{120} - \cdots \]
The even-power series with leading one is \( \cos x \), and the all-positive version is \( \sinh x \)
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In the Taylor expansion of \( f \) about \( x = a \), the coefficient of \( (x-a)^{n} \) is
Answer: C
Writing \( f(x) = \sum c_{n}(x-a)^{n} \) and differentiating \( n \) times at \( x = a \) leaves only the \( n \)th term, giving \( f^{(n)}(a) = n!\,c_{n} \) and so
\[ c_{n} \;=\; \frac{f^{(n)}(a)}{n!}. \]
The factorial divisor is what makes the series converge for well-behaved functions; omitting it produces a series that diverges for all but the smallest arguments
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Using two terms of its Maclaurin series, \( \sqrt{1.02} \) is approximately
Answer: B
The binomial expansion for small \( t \) is \( (1+t)^{1/2} \approx 1 + t/2 \). With \( t = 0.02 \),
\[ \sqrt{1.02} \;\approx\; 1 + \frac{0.02}{2} \;=\; 1.01. \]
The next term is \( -t^{2}/8 = -0.00005 \), so the true value is 1.00995 and the two-term estimate is high by five parts in a million
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The Maclaurin series for \( \ln(1+x) \) is valid for
Answer: A
The series \( x - x^{2}/2 + x^{3}/3 - \cdots \) has radius of convergence one. At \( x = 1 \) it becomes the alternating harmonic series, which converges to \( \ln 2 \); at \( x = -1 \) it is the harmonic series with a minus sign and diverges, matching \( \ln 0 \). So the interval is closed at the right and open at the left
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L'Hopital's rule may be applied directly to \( \lim_{x \to a} f(x)/g(x) \) when the quotient takes the form
Answer: D
The rule replaces the quotient by \( f'/g' \) and requires the original to be \( 0/0 \) or \( \infty/\infty \). The other indeterminate forms must first be rearranged into one of those two, a product by inverting one factor and a power by taking logarithms. Applying it to a determinate quotient gives a wrong answer
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The value of \( \lim_{x \to 0} x \ln x \) is
Answer: C
The form is \( 0 \times (-\infty) \), so rewrite it as a quotient and differentiate:
\[ \lim_{x \to 0^{+}} \frac{\ln x}{1/x} \;=\; \lim_{x \to 0^{+}} \frac{1/x}{-1/x^{2}} \;=\; \lim_{x \to 0^{+}} (-x) \;=\; 0. \]
The logarithm diverges, but only as slowly as a logarithm, and the factor \( x \) goes to zero faster than it grows
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The value of \( \lim_{x \to 0} (\cos x)^{1/x^{2}} \) is
Answer: B
The form is \( 1^{\infty} \), so take logarithms:
\[ \ln L \;=\; \lim_{x \to 0} \frac{\ln \cos x}{x^{2}} \;=\; \lim_{x \to 0} \frac{-x^{2}/2}{x^{2}} \;=\; -\frac{1}{2}, \]
using \( \ln \cos x \approx -x^{2}/2 \). Hence \( L = e^{-1/2} \). Reading \( 1^{\infty} \) as one is the standard error, the form being indeterminate precisely because the base approaches one at a rate that matters
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Put these steps of evaluating an indeterminate power in order. 1. Exponentiate the limit of the logarithm 2. Recognise the form as \( 1^{\infty} \), \( 0^{0} \) or \( \infty^{0} \) 3. Evaluate the resulting quotient, by L'Hopital if needed 4. Take logarithms to convert the power into a product, then a quotient
Answer: A
The form must be recognised before anything is done to it, since a determinate power needs no such treatment. Logarithms then turn the power into \( g \ln f \), a product that becomes a quotient. That quotient is evaluated, and only at the end is the exponential taken to recover the limit itself rather than its logarithm
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The \( n \)th derivative of \( e^{ax} \) is
Answer: D
Each differentiation multiplies by the constant \( a \) and leaves the exponential alone, so after \( n \) steps
\[ \frac{d^{n}}{dx^{n}} e^{ax} \;=\; a^{n} e^{ax}. \]
This is why the exponential is the natural basis for linear differential equations: substituting it turns each derivative into a power of \( a \) and the equation into a polynomial
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Leibniz's rule for the \( n \)th derivative of a product \( uv \) reads
Answer: C
Repeated use of the product rule generates binomial coefficients, exactly as in the binomial theorem:
\[ (uv)^{(n)} \;=\; \sum_{r=0}^{n} \binom{n}{r} u^{(n-r)} v^{(r)}. \]
For \( n = 1 \) this is \( u'v + uv' \). The two-term forms offered omit every mixed derivative between the extremes, which for \( n = 2 \) loses the \( 2u'v' \) term
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