Question

Math

Posted 5 months ago

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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 .
```

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Answer from Sia

Posted 5 months ago

Solution by Steps

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 $\lim_{x \to 0} g(x) = \lim_{x \to 0} h(x) = L$, then $\lim_{x \to 0} f(x) = L$

step 3

First, we check the limits of $g(x)$ and $h(x)$ as $x$ approaches $0$

step 4

$\lim_{x \to 0} g(x) = \lim_{x \to 0} (-x^2 + 0.5) = 0.5$

step 5

$\lim_{x \to 0} h(x) = \lim_{x \to 0} (x^2 + 1.5) = 1.5$

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.

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