We want to find $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}$. Direct substitution and other algebraic methods don't seem to work.

Looking at the graph of $f(x)=\frac{x}{\sin (x)}$, we can estimate that the limit is equal to 1 .

To prove that $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}=1$, we can use the squeeze theorem.
Pablo suggested that we use the functions $g(x)=-x^{2}+0.5$ and $h(x)=x^{2}+1.5$ in order to apply the squeeze theorem.

Does Pablo's suggestion seem to be correct?
Choose 1 answer:
(A) Yes, Pablo's suggestion seems to be correct.
(B) No, Pablo's suggestion is incorrect because it's not true that one function is always below $f$ and one function is always above it for $x$-values near 0 .
(c) No, Pablo's suggestion is incorrect because it's not true that the limits of $g$ and $h$ are both equal to 1 .

Q: We want to find $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}$. Direct substitution and other algebraic methods don't seem to work. Looking at the graph of $f(x)=\frac{x}{\sin (x)}$, we can estimate that the limit is equal to 1 . To prove that $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}=1$, we can use the squeeze theorem. Pablo suggested that we use the functions $g(x)=-x^{2}+0.5$ and $h(x)=x^{2}+1.5$ in order to apply the squeeze theorem. Does Pablo's suggestion seem to be correct? Choose 1 answer: (A) Yes, Pablo's suggestion seems to be correct. (B) No, Pablo's suggestion is incorrect because it's not true that one function is always below $f$ and one function is always above it for $x$-values near 0 . (c) No, Pablo's suggestion is incorrect because it's not true that the limits of $g$ and $h$ are both equal to 1 .

∻Solution by Steps∻ ‖ step 1 ⋮ We need to find $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}$. Direct substitution and other algebraic methods don't seem to work. ‖ ‖ step 2 ⋮ Looking at the graph of $f(x)=\frac{x}{\sin (x)}$, we can estimate that the limit is equal to 1. ‖ ‖ step 3 ⋮ To prove that $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}=1$, we can use the squeeze theorem. ‖ ‖ step 4 ⋮ Pablo suggested that we use the functions $g(x)=-x^{2}+0.5$ and $h(x)=x^{2}+1.5$ in order to apply the squeeze theorem. ‖ ‖ step 5 ⋮ The limits of $g(x)$ and $h(x)$ as $x$ approaches 0 are $\lim_{x \to 0} g(x) = 0.5$ and $\lim_{x \to 0} h(x) = 1.5$. ‖ ‖ step 6 ⋮ Since the limits of $g(x)$ and $h(x)$ are not both equal to 1, Pablo's suggestion is incorrect. ‖ ***C*** ∻Key Concept∻ ⚹Squeeze Theorem⚹ ∻Explanation∻ ⚹The squeeze theorem requires that the limits of the bounding functions be equal to the limit of the function in question. Since the limits of $g(x)$ and $h(x)$ are not both equal to 1, Pablo's suggestion is incorrect.⚹◊What is the formula for the squeeze theorem used to determine limits of functions?⍭ Generate me a similar question◊

Question

We want to find $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}$. Direct substitution and other algebraic methods don't seem to work.

Looking at the graph of $f(x)=\frac{x}{\sin (x)}$, we can estimate that the limit is equal to 1 .

To prove that $\lim _{x \rightarrow 0} \frac{x}{\sin (x)}=1$, we can use the squeeze theorem.
Pablo suggested that we use the functions $g(x)=-x^{2}+0.5$ and $h(x)=x^{2}+1.5$ in order to apply the squeeze theorem.

Does Pablo's suggestion seem to be correct?
Choose 1 answer:
(A) Yes, Pablo's suggestion seems to be correct.
(B) No, Pablo's suggestion is incorrect because it's not true that one function is always below $f$ and one function is always above it for $x$-values near 0 .
(c) No, Pablo's suggestion is incorrect because it's not true that the limits of $g$ and $h$ are both equal to 1 .

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We need to find $\lim _{x ightarrow 0} \frac{x}{\sin (x)}$. Direct substitution and other algebraic methods don't seem to work. ‖ ‖ step 2 ⋮ Looking at the graph of $f(x)=\frac{x}{\sin (x)}$, we can estimate that the limit is equal to 1. ‖ ‖ step 3 ⋮ To prove that $\lim _{x ightarrow 0} \frac{x}{\sin (x)}=1$, we can use the squeeze theorem. ‖ ‖ step 4 ⋮ Pablo suggested that we use the functions $g(x)=-x^{2}+0.5$ and $h(x)=x^{2}+1.5$ in order to apply the squeeze theorem. ‖ ‖ step 5 ⋮ The limits of $g(x)$ and $h(x)$ as $x$ approaches 0 are $\lim_{x o 0} g(x) = 0.5$ and $\lim_{x o 0} h(x) = 1.5$. ‖ ‖ step 6 ⋮ Since the limits of $g(x)$ and $h(x)$ are not both equal to 1, Pablo's suggestion is incorrect. ‖ ***C*** ∻Key Concept∻ ⚹Squeeze Theorem⚹ ∻Explanation∻ ⚹The squeeze theorem requires that the limits of the bounding functions be equal to the limit of the function in question. Since the limits of $g(x)$ and $h(x)$ are not both equal to 1, Pablo's suggestion is incorrect.⚹◊What is the formula for the squeeze theorem used to determine limits of functions?⍭ Generate me a similar question◊

Sia