Find lim_(x rarr0)(sin(x))/(sin(2x)). Choose 1 answer: (A) (1)/(2) (B) 1 (C) 2 (D) The limit doesn't exist

Apply Limit Directly: We need to find the limit of the function sin(x)/sin(2x) as x approaches 0. We can start by applying the limit to the function directly.Identify Indeterminate Form: We notice that if we directly substitute x = 0 into the function, we get sin(0)/sin(0), which is an indeterminate form 0/0. This means we cannot directly evaluate the limit by substitution.Use Trigonometric Identity: To resolve the indeterminate form, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x) to rewrite the denominator. lim_(x -> 0)(sin(x))/(sin(2x)) = lim_(x -> 0)(sin(x))/(2sin(x)cos(x))Simplify Expression: We can now simplify the expression by canceling out the common sin(x) term in the numerator and denominator, as long as x is not equal to 0 (since sin(0) = 0, we cannot divide by zero). lim_(x -> 0)(sin(x))/(2sin(x)cos(x)) = lim_(x -> 0)(1)/(2cos(x))Substitute x = 0: Now that we have simplified the expression, we can directly substitute x = 0 into the remaining function, as cos(x) is continuous at x = 0. lim_(x -> 0)(1)/(2cos(x)) = (1)/(2cos(0)) = (1)/(2*1) = 1/2

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Find
lim_(x rarr0)(sin(x))/(sin(2x)).
Choose 1 answer:
(A)
(1)/(2)
(B) 1
(C) 2
(D) The limit doesn't exist

Find $\lim _{x \rightarrow 0} \frac{\sin (x)}{\sin (2 x)}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{1}{2}$ $\newline$ (B) $1$ $\newline$ (C) $2$ $\newline$ (D) The limit doesn't exist

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Calculus

Find derivatives of logarithmic functions

Full solution

Q. Find

\lim _{x \rightarrow 0} \frac{\sin (x)}{\sin (2 x)}

\newline

Choose

1

answer:

\newline

(A)

\frac{1}{2}

\newline

(B)

1

\newline

(C)

2

\newline

(D) The limit doesn't exist

Apply Limit Directly: We need to find the limit of the function $\frac{\sin(x)}{\sin(2x)}$ as $x$ approaches $0$ . We can start by applying the limit to the function directly.
Identify Indeterminate Form: We notice that if we directly substitute $x = 0$ into the function, we get $\frac{\sin(0)}{\sin(0)}$ , which is an indeterminate form $\frac{0}{0}$ . This means we cannot directly evaluate the limit by substitution.
Use Trigonometric Identity: To resolve the indeterminate form, we can use the trigonometric identity $\sin(2x) = 2\sin(x)\cos(x)$ to rewrite the denominator. $\newline$ $\lim_{x \to 0}\frac{\sin(x)}{\sin(2x)} = \lim_{x \to 0}\frac{\sin(x)}{2\sin(x)\cos(x)}$
Simplify Expression: We can now simplify the expression by canceling out the common $\sin(x)$ term in the numerator and denominator, as long as $x$ is not equal to $0$ (since $\sin(0) = 0$ , we cannot divide by zero). $\newline$ $\lim_{x \to 0}\frac{\sin(x)}{2\sin(x)\cos(x)} = \lim_{x \to 0}\frac{1}{2\cos(x)}$
Substitute $x = 0$ : Now that we have simplified the expression, we can directly substitute $x = 0$ into the remaining function, as $\cos(x)$ is continuous at $x = 0$ . $\lim_{x \to 0}\frac{1}{2\cos(x)} = \frac{1}{2\cos(0)} = \frac{1}{2\cdot 1} = \frac{1}{2}$