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\[\large \lim_{x\to 0} \frac{x^{2}}{\cos(5x)-\tan^{2}(3x)-1}\] Evaluate the limit above without using L'Hôpital rule.

Note by Majed Musleh 2 years, 2 months ago

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\( \large \lim_{x\to 0} \dfrac{x^{2}}{\cos(5x)-\tan^{2}(3x)-1}\)

\( = \large \lim_{x\to 0} \dfrac{1}{\frac{\cos(5x)- 1}{x^2} - \frac{\tan^{2}(3x)}{x^2}} \)

\( = \large \lim_{x\to 0} \dfrac{1}{25\frac{\cos(5x)- 1}{(5x)^2} - 9\frac{\tan^{2}(3x)}{(3x)^2}} \)

\( = \dfrac{1}{\frac{-25}{2} - 9} \)

\( = \frac{ -2}{43} \)

I only use \( \large \lim_{x\to 0} \frac{\tan(x)}{x} = 1 \) and \( \large \lim_{x\to 0} \frac{\cos(x) - 1}{x^2} = \frac{-1}{2} \)

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One way would be to use the relevant Maclaurin series for the cosine and tangent functions. The limit would then become

\(\lim_{x \rightarrow 0} \dfrac{x^{2}}{(1 - \dfrac{(5x)^{2}}{2} + O(x^{4})) - ((3x) + O(x^{3}))^{2} - 1} =\)

\(\lim_{x \rightarrow 0} \dfrac{x^{2}}{-\dfrac{25x^{2}}{2} - 9x^{2} + O(x^{4})} = \lim_{x \rightarrow 0} \dfrac{1}{-\dfrac{25}{2} - 9 + O(x^{2})} = -\dfrac{2}{43}.\)

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## Comments

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TopNewest\( \large \lim_{x\to 0} \dfrac{x^{2}}{\cos(5x)-\tan^{2}(3x)-1}\)

\( = \large \lim_{x\to 0} \dfrac{1}{\frac{\cos(5x)- 1}{x^2} - \frac{\tan^{2}(3x)}{x^2}} \)

\( = \large \lim_{x\to 0} \dfrac{1}{25\frac{\cos(5x)- 1}{(5x)^2} - 9\frac{\tan^{2}(3x)}{(3x)^2}} \)

\( = \dfrac{1}{\frac{-25}{2} - 9} \)

\( = \frac{ -2}{43} \)

I only use \( \large \lim_{x\to 0} \frac{\tan(x)}{x} = 1 \) and \( \large \lim_{x\to 0} \frac{\cos(x) - 1}{x^2} = \frac{-1}{2} \)

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One way would be to use the relevant Maclaurin series for the cosine and tangent functions. The limit would then become

\(\lim_{x \rightarrow 0} \dfrac{x^{2}}{(1 - \dfrac{(5x)^{2}}{2} + O(x^{4})) - ((3x) + O(x^{3}))^{2} - 1} =\)

\(\lim_{x \rightarrow 0} \dfrac{x^{2}}{-\dfrac{25x^{2}}{2} - 9x^{2} + O(x^{4})} = \lim_{x \rightarrow 0} \dfrac{1}{-\dfrac{25}{2} - 9 + O(x^{2})} = -\dfrac{2}{43}.\)

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