CALCU- _x x+ (2 x)+1x^2-^2 is(A) 12 (B) 1(C) 1(

\lim _{x \rightarrow \pi} \frac{\cos x+\sin (2 x)+1}{x^{2}-\pi^{2}}

is (A)

\frac{1}{2 \pi}

(B)

\frac{1}{\pi}

(C) 1 (D) nonexistent

D

Key Concept

L'Hôpital's Rule and Indeterminate Forms

Explanation

The given limit is of the indeterminate form

\frac{0}{0}

as

x

approaches

\pi

. To evaluate it, we can apply L'Hôpital's Rule, which states that if the limit

\lim_{x \to a} \frac{f(x)}{g(x)}

results in an indeterminate form

\frac{0}{0}

or

\frac{\infty}{\infty}

, then it can be computed as

\lim_{x \to a} \frac{f'(x)}{g'(x)}

, provided that the derivatives exist and the new limit exists or is

\infty

. Differentiating the numerator and denominator separately and then taking the limit should give us the answer. However, in this case, the limit does not exist because after applying L'Hôpital's Rule, the resulting limit still leads to an indeterminate form that does not resolve to a finite number or infinity.

If

f(x)=\int_{1}^{x^{3}} \frac{1}{1+\ln t} d t

for

x \geq 1

, then

f^{\prime}(2)=

(A)

\frac{1}{1+\ln 2}

(B)

\frac{12}{1+\ln 2}

(C)

\frac{1}{1+\ln 8}

(D)

\frac{12}{1+\ln 8}

D

Key Concept

Fundamental Theorem of Calculus and Chain Rule

Explanation

To find

f'(x)

, we apply the Fundamental Theorem of Calculus and then use the Chain Rule to differentiate the upper limit of the integral. Specifically,

f'(x) = \\frac{d}{dx} \\int_{1}^{x^{3}} \\frac{1}{1+\\ln t} dt = \\frac{1}{1+\\ln(x^3)} \\cdot \\frac{d}{dx}(x^3)

. Evaluating at

x=2

gives

f'(2) = \\frac{1}{1+\\ln(8)} \\cdot 3 \\cdot 2^2 = \\frac{12}{1+\\ln 8}

.

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