Max tried to find $\lim _{x \rightarrow \frac{\pi}{2}} \frac{\sin (2 x)}{\cos (x)}$ . $\newline$ Using direct substitution, he got $\frac{0}{0}$ . $\newline$ For Max's next step, which method would apply? $\newline$ Choose $1$ answer: $\newline$ (A) Factorization and cancellation $\newline$ (B) Rationalization using conjugates $\newline$ (C) Alternate forms of trigonometric functions

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Algebra 2

Symmetry and periodicity of trigonometric functions

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Q. Max tried to find

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

\newline

Using direct substitution, he got

\frac{0}{0}

\newline

For Max's next step, which method would apply?

\newline

Choose

1

answer:

\newline

(A) Factorization and cancellation

\newline

(B) Rationalization using conjugates

\newline

Identify Indeterminate Form: Max encountered an indeterminate form $0/0$ when he directly substituted $x = \frac{\pi}{2}$ into the limit expression $\frac{\sin(2x)}{\cos(x)}$ . To resolve this, he needs to apply a method that simplifies the expression and potentially eliminates the indeterminate form. Let's analyze the given choices.
Analyze Given Choices: Option (A) suggests factorization and cancellation. This method is useful when the numerator and denominator can be factored, and common factors can be canceled out. However, in the expression $(\sin(2x))/(\cos(x))$ , there are no obvious factors that can be canceled directly. Therefore, this method does not seem applicable.
Factorization Not Applicable: Option (B) suggests rationalization using conjugates. This method is typically used for limits involving square roots where multiplying by the conjugate can help eliminate the square roots. Since there are no square roots in the expression $(\sin(2x))/(\cos(x))$ , this method is not suitable.
Rationalization Not Suitable: Option (C) suggests using alternate forms of trigonometric functions. This method involves using trigonometric identities to rewrite the expression in a form that may resolve the indeterminate form. For example, we can use the double-angle identity for sine, which states that $\sin(2x) = 2\sin(x)\cos(x)$ . This could potentially simplify the expression and help us evaluate the limit.
Use Trigonometric Identities: Let's apply the double-angle identity to the expression $(\sin(2x))/(\cos(x))$ and see if it simplifies: $\newline$ $(\sin(2x))/(\cos(x)) = (2\sin(x)\cos(x))/(\cos(x))$ $\newline$ Now, we can cancel the common factor of $\cos(x)$ in the numerator and denominator, provided that $\cos(x)$ is not zero. Since we are considering the limit as $x$ approaches $\pi/2$ , and $\cos(\pi/2) = 0$ , we need to be careful not to cancel out a term that is zero. However, since we are taking the limit, we are not evaluating at $x = \pi/2$ exactly, but rather approaching it, so $\cos(x)$ is not exactly zero in the neighborhood around $\pi/2$ . $\newline$ After cancellation, we get: $\newline$ $(\sin(2x))/(\cos(x)) = (2\sin(x)\cos(x))/(\cos(x))$ $0$ $\newline$ Now we can evaluate the limit as $x$ approaches $\pi/2$ : $\newline$ $(\sin(2x))/(\cos(x)) = (2\sin(x)\cos(x))/(\cos(x))$ $3$