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lutik1710 [3]
1 year ago
9

Evaluate : limx→ tan2-sin2x x3

Mathematics
1 answer:
GrogVix [38]1 year ago
8 0

Given:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Solve:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Use l'hopital's rule:

\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}

Simplify:

\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}

Apply the constant multiple rule:

\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}

Similary :

\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}

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You should first turn the equation into y intercept form:
7y - 4x = -14; Add 4x from both sides
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y = 4/7x - 2

Since you did not include choices, the answer will be the line with the slope 4/7 and without the y-intercept -2, because lines that are parallel will always have the same slope.

Example Answers: y = 4/7x + 2, y= 4/7x - 4
y = 4/7x + 10

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

The problem wants you to PLUG IN x into the correct equation.

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f(3) means 3 equals x

For example g(9) means 9 = x and f(3/5) means 3/5 = x and so on.

2) Now that you know x equals two numbers -1 and 3 you have to plug them into a equation.

3) Which equation do you plug it in you may ask? You must plug in -1 into x^2 +3 BECAUSE the condition next to it says you can only plug in numbers to x^2+3 greater than or equal to -1 and less than 2. Therefore, you can plug in -1 to this equation.

-1^2 +3 = <u>4</u>

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