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AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 89 of 160

Marks: +1, -0

\lim _{x \rightarrow 0} \frac{(1-\cos 2 x) \cdot \sin 5 x}{x^2 \sin 3 x}

Solution:

Concept:

Use the standard limit

\displaystyle\lim_{\theta\to0}\frac{\sin\theta}{\theta}=1

and the identity

1-\cos2x=2\sin^2x

to simplify the expression.

Explanation:

Start with the given limit:

\displaystyle\lim_{x\to0}\frac{(1-\cos2x)\sin5x}{x^2\sin3x}

Replace

1-\cos2x

2\sin^2x

\displaystyle\lim_{x\to0}\frac{2\sin^2x\cdot\sin5x}{x^2\sin3x}

Rewrite the numerator and denominator to match the standard forms:

\displaystyle\lim_{x\to0}\frac{2\sin^2x}{x^2}\cdot\frac{\sin5x}{x}\cdot\frac{x}{\sin3x}

Now use the limit properties. Let

\displaystyle L=\lim_{x\to0}\frac{\sin x}{x}=1

. Then:

\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=L^2=1

\displaystyle\lim_{x\to0}\frac{\sin5x}{x}=5\cdot\lim_{x\to0}\frac{\sin5x}{5x}=5\cdot1=5

\displaystyle\lim_{x\to0}\frac{x}{\sin3x}=\frac{1}{\displaystyle\lim_{x\to0}\frac{\sin3x}{3x}}\cdot\frac{1}{3}=\frac{1}{1}\cdot\frac{1}{3}=\frac13

Multiply all parts together with the factor 2:

2\times1\times5\times\frac13=\frac{10}{3}

Answer:

The value of the limit is

\frac{10}{3}

, which corresponds to Option A.

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