Limit

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

#Calculus

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Comments

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

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

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

Brian Charlesworth - 5 years, 8 months ago

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

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

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

$= \dfrac{1}{\frac{-25}{2} - 9}$

$= \frac{ -2}{43}$

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

Siddhartha Srivastava - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$