evaluate the following: $lim_{x\rightarrow0}\frac{sin x}{x}$\n1\n$infty$\n-1\ndne\n0\n$-infty$\nquestion…

evaluate the following: $lim_{x\rightarrow0}\frac{sin x}{x}$\n1\n$infty$\n-1\ndne\n0\n$-infty$\nquestion 11\n1 pts\nsuppose that on the interval $(-2,5)$, we know that $4 - x^{2}leq f(x)leq5 - 2x$. what is $lim_{x\rightarrow1}f(x)$?\n-2\n0\ndne\n5\n3

evaluate the following: $lim_{x\rightarrow0}\frac{sin x}{x}$\n1\n$infty$\n-1\ndne\n0\n$-infty$\nquestion 11\n1 pts\nsuppose that on the interval $(-2,5)$, we know that $4 - x^{2}leq f(x)leq5 - 2x$. what is $lim_{x\rightarrow1}f(x)$?\n-2\n0\ndne\n5\n3

Answer

Explanation:

Step1: Recall important limit

The limit $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$ is a well - known limit in calculus. It can be proven geometrically or using L'Hopital's rule.

Step2: For the second limit, use the Squeeze Theorem

The Squeeze Theorem states that if $g(x)\leq f(x)\leq h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $x = a$) and $\lim_{x\rightarrow a}g(x)=\lim_{x\rightarrow a}h(x)=L$, then $\lim_{x\rightarrow a}f(x)=L$. First, find $\lim_{x\rightarrow1}(4 - x^{2})$: Substitute $x = 1$ into $4 - x^{2}$, we get $4-1^{2}=3$. Then, find $\lim_{x\rightarrow1}(5 - 2x)$: Substitute $x = 1$ into $5 - 2x$, we get $5-2\times1 = 3$. Since $\lim_{x\rightarrow1}(4 - x^{2})=\lim_{x\rightarrow1}(5 - 2x)=3$ and $4 - x^{2}\leq f(x)\leq5 - 2x$, by the Squeeze Theorem, $\lim_{x\rightarrow1}f(x)=3$.

Answer:

  1. A. 1
  2. E. 3