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 .

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 determine if Pablo's suggestion to use the functions $g(x) = -x^2 + 0.5$ and $h(x) = x^2 + 1.5$ is correct for applying the squeeze theorem to find $\lim_{x o 0} \frac{x}{\sin(x)}$. ‖ ‖ step 2 ⋮ The squeeze theorem states that if $g(x) \leq f(x) \leq h(x)$ for all $x$ in some interval around $0$ (except possibly at $0$ itself), and $\lim_{x o 0} g(x) = \lim_{x o 0} h(x) = L$, then $\lim_{x o 0} f(x) = L$. ‖ ‖ step 3 ⋮ First, we check the limits of $g(x)$ and $h(x)$ as $x$ approaches $0$. ‖ ‖ step 4 ⋮ $\lim_{x o 0} g(x) = \lim_{x o 0} (-x^2 + 0.5) = 0.5$ ‖ ‖ step 5 ⋮ $\lim_{x o 0} h(x) = \lim_{x o 0} (x^2 + 1.5) = 1.5$ ‖ ‖ step 6 ⋮ Since the limits of $g(x)$ and $h(x)$ are not 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 equal to $1$, they cannot be used to apply the squeeze theorem in this case.⚹◊What is the significance of using the squeeze theorem in finding limits of functions?⍭ Generate me a similar question◊

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