Calculus

LIMIT X TENDS TO 0 SIN5X-SIN3X/SINX???

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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}`	$\boxed{123}$

Comments

Hi, rishit, firstly, you should try to learn latex here

Different Interpretations:

$1. \displaystyle \lim_{x \to 0} \sin 5x - \sin \bigg(\frac{3x}{\sin x}\bigg) = -\sin 3$

$2.\displaystyle \lim _{x \to 0} \sin 5x - \frac{\sin 3x}{\sin x} = -3$

$3. \displaystyle \lim_{x \to 0} \frac{\sin 5x - \sin 3x}{\sin x} = \lim_{x \to 0} \frac{2 \sin x \cos4x}{\sin x} = 2$

jatin yadav - 7 years, 6 months ago

Using L' Hopital's Rule, we take the derivative of the numerator and denominator of the function so we have (5 cos(5x) - 3 cos(3x)) / cos x, this is the time to take the limit and simply substitute the value... so this limit as the function tends to zero is 2.

John Ashley Capellan - 7 years, 6 months ago

-3

Vinayak Bodake - 7 years, 6 months ago

Can you explain your thinking step by step?

Calvin Lin Staff - 7 years, 6 months ago

Obviously this person considers the expression as $\sin 5x - \frac{\sin 3x}{\sin x}$ , which is a perfectly valid interpretation. In fact, the one with $\frac{\sin 5x - \sin 3x}{\sin x}$ is wrong since it would be (sin 5x - sin 3x)/sin x in linear form; note the parentheses.

(Moreover, I believe the correct interpretation according to how the problem is currently phrased is thus $\sin 5x - \sin \frac{3x}{\sin x}$ . In this case it seems that the limit is $-\sin 3$ .)

Ivan Koswara - 7 years, 6 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}`	$\boxed{123}$