B

1003109902

Level: 
B
Find the next possible correct step in evaluating this limit. \[ \lim\limits_{x\to\infty}\frac{\sqrt{x^2+1}-x}{x+1} \]
\( \lim\limits_{x\to\infty}\frac{\sqrt{1+\frac1{x^2}}-1}{1+\frac1x} \)
\( \lim\limits_{x\to\infty}\frac{\sqrt{1+\frac1{x^2}}-\frac1x}{1+\frac1x} \)
\( \lim\limits_{x\to\infty}\frac{\sqrt{x+\frac1x}-1}{1+\frac1x} \)
\( \lim\limits_{x\to\infty}\frac{\sqrt{1+\frac1{x^2}}-\frac1x}{\frac1x+\frac1{x^2} } \)

1003109901

Level: 
B
Find the next possible correct step in evaluating this limit. \[ \lim\limits_{x\to\infty}\frac{2x-1}{\sqrt{2x^2-1}} \]
\( \lim\limits_{x\to\infty}⁡\frac{2-\frac1x}{\sqrt{2-\frac1{x^2}}} \)
\( \lim\limits_{x\to\infty}\frac{2-\frac1x}{\sqrt{2x-\frac1x}} \)
\( \lim\limits_{x\to\infty}\frac{2-\frac1x}{\sqrt{2-x}} \)
\( \lim\limits_{x\to\infty}\frac{\frac2x-\frac1{x^2}}{\sqrt{2-\frac1{x^2}}} \)

1003108205

Level: 
B
Compare the two definite integrals \( I_1=\int\limits_{-1}^1\left(x+\frac{\pi}2\right)\mathrm{d}x \) and \( I_2=\int\limits_0^{\frac{\pi}4}\mathrm{tg}\,x\cdot\cos ⁡x\,\mathrm{d}x \).
\( I_1 \) is bigger than \( I_2 \).
\( I_1 \) is smaller than \( I_2 \).
\( I_1 \) is equal to \( I_2 \).
These integrals cannot be compared.

1003108203

Level: 
B
Compare the definite integral \( I=\int\limits_0^{\frac{\pi}4}\frac{\cos⁡2b}{\cos^2⁡b}\,\mathrm{d}b \) to the number \( \frac{\pi}2 \).
\( I \) is smaller than \( \frac{\pi}2 \) by \( 1 \).
\( I \) is bigger than \( \frac{\pi}2 \) by \( 1 \).
\( I \) is equal to \( \frac{\pi}2 \).
\( I \) is smaller than \( \frac{\pi}2 \) by \( \frac{\pi}4 \).