Suppose $-x^2-9\le v(x)\le 4x-5$ for all $x$ near $-2$. What is ${\lim }_{x\to -2}v(x)$?

Evaluate ${\lim }_{x\to 0}{x}^4\cos \left(\frac{6}{x}\right)$, with the angle in radians.

Evaluate ${\lim }_{x\to 0}{x}^4\sin \left(\frac{2}{x}\right)$, with the angle in radians.

Suppose $-x^2-2\le u(x)\le 2x-1$ for all $x$ near $-1$. What is ${\lim }_{x\to -1}u(x)$?

A student argues: since $\sin \left(\frac{2}{x}\right)$ always lies between $-1$ and $1$, it is bounded, so ${\lim }_{x\to 0}\sin \left(\frac{2}{x}\right)=0$. Is the argument valid?

Suppose $x+1\le g(x)\le x^2+2$ for all $x$ near $-3$. What can be concluded about ${\lim }_{x\to -3}g(x)$?

Evaluate ${\lim }_{x\to 0}{x}^5\cos \left(\frac{2}{x}\right)$, with the angle in radians.

Evaluate ${\lim }_{x\to 0}{x}^4\cos \left(\frac{3}{x}\right)$, with the angle in radians.

Suppose $24-x^2\le f(x)\le 40-8x$ for all $x$ near $4$. What is ${\lim }_{x\to 4}f(x)$?

A student argues: since $\sin \left(\frac{4}{x}\right)$ always lies between $-1$ and $1$, it is bounded, so ${\lim }_{x\to 0}\sin \left(\frac{4}{x}\right)=0$. Is the argument valid?

Which inequality is the correct starting point for evaluating ${\lim }_{x\to 0}{x}^6\cos \left(\frac{3}{x}\right)$ by the Squeeze Theorem?

Evaluate ${\lim }_{x\to 0}{x}^5\sin \left(\frac{3}{x}\right)$, with the angle in radians.

Suppose $-x^2-14\le u(x)\le 4x-10$ for all $x$ near $-2$. What is ${\lim }_{x\to -2}u(x)$?

Which inequality is the correct starting point for evaluating ${\lim }_{x\to 0}{x}^5\cos \left(\frac{6}{x}\right)$ by the Squeeze Theorem?

Suppose $x+3\le p(x)\le x^2+2$ for all $x$ near $3$. What can be concluded about ${\lim }_{x\to 3}p(x)$?

Suppose $-x^2-9\le v(x)\le -4x-5$ for all $x$ near $2$. What is ${\lim }_{x\to 2}v(x)$?

Evaluate ${\lim }_{x\to 0}{x}^3\cos \left(\frac{4}{x}\right)$, with the angle in radians.

Suppose $x+4\le g(x)\le x^2-2$ for all $x$ near $4$. What can be concluded about ${\lim }_{x\to 4}g(x)$?

Which inequality is the correct starting point for evaluating ${\lim }_{x\to 0}{x}^2\cos \left(\frac{3}{x}\right)$ by the Squeeze Theorem?

A student argues: since $\sin \left(\frac{8}{x}\right)$ always lies between $-1$ and $1$, it is bounded, so ${\lim }_{x\to 0}\sin \left(\frac{8}{x}\right)=0$. Is the argument valid?