1003163907 Level: AEvaluate the following limit. (Apply L'Hospital's rule repeatedly.) \[ \lim_{x\to\infty}\frac{\mathrm{e}^x-2}{x^2} \]\( \infty \)\( 0 \)\( 1 \)\( \frac12 \)
1003163906 Level: AEvaluate the following limit. (Apply L'Hospital's rule repeatedly.) \[ \lim_{x\to\infty}\frac{x^2-1}{x^3-2x^2+x} \]\( 0 \)\( \infty \)\( \frac23 \)\( 1 \)
1003163905 Level: AUse L'Hospital's rule to find the following limit. \[ \lim_{x\to1}\frac{\sqrt{x+3}-2}{\ln x} \]\( \frac14 \)\( \frac12 \)\( 0 \)\( 1 \)\( \frac{\sqrt2}2 \)
1003163904 Level: AUse L'Hospital's rule to find the following limit. \[ \lim_{x\to\frac{\pi}3}\frac{1-2\cos x}{\pi-3x} \]\( -\frac{\sqrt3}3 \)\( -\frac{\sqrt3}6 \)\( \frac{\sqrt3}3 \)\( \frac{\sqrt3}6 \)\( -\frac13 \)
1003163903 Level: AUse L'Hospital's rule to find the following limit. \[ \lim_{x\to0}\frac{\sin2x}{\mathrm{tg}\,3x} \]\( \frac23 \)\( 1 \)\( 2 \)\( 0 \)\( 6 \)
1003163902 Level: AUse L'Hospital's rule to find the following limit. \[ \lim_{x\to0}\frac{\mathrm{e}^x-1}{\sin2x} \]\( \frac12 \)\( -\frac12 \)\( 1 \)\( 0 \)\( -1 \)
1003163901 Level: AUse L'Hospital's rule to find the following limit. \[ \lim_{x\to2}\frac{2x^3-3x^2-4}{x^2+x-6} \]\( \frac{12}5 \)\( \frac{18}5 \)\( \frac{12}3 \)\( 0 \)\( \infty \)
9000145401 Level: CIdentify a true statement on the function \(f(x) = 2x^{3} + 3x^{2} - 12x - 12\).The function \(f\) has a local maximum at the point \(x = -2\).The function \(f\) has a local minimum at the point \(x = -2\)..The global maximum of \(f\) on \(\mathbb{R}\) is at \(x = -2\).The global minimum of \(f\) on \(\mathbb{R}\) is at \(x = -2\).
9000145402 Level: CIdentify a true statement about the function \(f(x) = 2x^{2} -\frac{x^{4}} {4} \).The global maximum of \(f\) on \(\mathbb{R}\) is at \(x = 2\) and \(x = -2\).The global minimum of \(f\) on \(\mathbb{R}\) is at \(x = 2\) and \(x = -2\).The function \(f\) has a local minimum at the point \(x = 2\).The function \(f\) has a local minimum at the point \(x = -2\).
9000145403 Level: CIdentify a true statement on the function \(f(x)= \frac{4-3x} {x\left (1-x\right )}\).The function \(f\) has a local minimum at the point \(x = \frac{2} {3}\).The function \(f\) has a local maximum at the point \(x = \frac{2} {3}\).The global maximum of \(f\) on \(\mathbb{R}\setminus \{0.1\}\) is at \(x = \frac{2} {3}\).The global minimum of \(f\) on \(\mathbb{R}\setminus \{0.1\}\) is at \(x = \frac{2} {3}\).