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AP DSC School Assistant Math Model Test 3

Section: Subject Specific

Question : 77 of 160

Marks: +1, -0

\lim_{x\rightarrow 0} \frac{(1-\cos 2x)(3+\cos x)}{x\tan 4x}

is equal to:

Solution:

Concept:

Use standard limit formulas:

\displaystyle\lim_{x\to 0}\frac{\sin x}{x}=1

and

\displaystyle\lim_{x\to 0}\frac{\tan x}{x}=1

. Also rewrite

1-\cos2x = 2\sin^2 x

and simplify the given expression accordingly.

Explanation:

Step 1: Write the given limit as:

\lim_{x\to0}\frac{1-\cos2x}{x\tan4x}\cdot(3+\cos x)

Step 2: Replace

1-\cos2x

with

2\sin^2 x

= \lim_{x\to0}\frac{2\sin^2 x}{x\tan4x}\cdot(3+\cos x)

Step 3: Separate the factors to use standard limits:

= \lim_{x\to0}2\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{x}\cdot\frac{1}{\frac{\tan4x}{4x}\cdot4}\cdot(3+\cos x)

Better write:

\displaystyle \frac{2\sin^2 x}{x\tan4x} = 2\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{4}\cdot\frac{4x}{\tan4x}

So the expression becomes:

\lim_{x\to0} 2\left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{4}\cdot\frac{4x}{\tan4x}\cdot(3+\cos x)

Step 4: Apply limits:

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

\displaystyle\lim_{x\to0}\frac{4x}{\tan4x}=1

, and

\displaystyle\lim_{x\to0}(3+\cos x)=3+1=4

. Therefore:

= 2 \times (1)^2 \times \frac{1}{4} \times 1 \times 4 = 2

Answer:

2

(Option C)

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