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\( \lim _ { x \rightarrow - \infty } \frac { 1 + \sin ^ { 3 } x } { \cos ^ { 2 } x } \)

Ask by Kelley Osborne. in South Africa
Jan 19,2025

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The limit does not exist because the expression grows without bound as \( x \) approaches negative infinity.

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Mind Expander

To evaluate the limit \( \lim _ { x \rightarrow - \infty } \frac { 1 + \sin ^ { 3 } x } { \cos ^ { 2 } x } \), we need to examine the behavior of the sine and cosine functions as \( x \) approaches negative infinity. 1. The sine function oscillates between -1 and 1 for all \( x \), so \( \sin^3 x \) will also oscillate between -1 and 1. Therefore, the expression \( 1 + \sin^3 x \) will oscillate between 0 and 2. 2. The cosine function also oscillates between -1 and 1, meaning \( \cos^2 x \) oscillates between 0 and 1. However, it will never be zero when considering the limit, since \( \cos^2 x \) is always non-negative. As such, we can deduce that: - In the worst scenario, when \( \cos^2 x \) is approaching 0, if \( 1 + \sin^3 x \) is not approaching 0, the fraction can become very large. - Conversely, when \( \cos^2 x \) is closer to 1, the fraction will be bounded between 0 and 2, depending on the oscillation of \( \sin^3 x \). However, since both \( 1 + \sin^3 x \) oscillates and \( \cos^2 x \) approaches zero at certain points, the limit does not settle on a specific value; instead, it oscillates wildly without converging. Thus, the limit does not exist.

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