Math, asked by anandmalgekar9054, 3 days ago

Lim x tends to 0 (sin 8x/tan 4x)

Answers

Answered by anindyaadhikari13
9

Solution:

Given Limit:

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) }  \bigg]

If we substitute x = 0 directly, we get:

 \displaystyle \rm =\dfrac{ \sin(0) }{ \tan(0) }

 \displaystyle \rm =\dfrac{0}{0}

Which is an indeterminate form.

Consider again:

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) }  \bigg]

We know that:

 \bigstar \: \underline{ \boxed{ \rm \sin(2x) = 2 \sin(x) \cos(x) }}

Using this result, we get:

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(2 \times 4x) }{ \tan(4x) }  \bigg]

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\sin(4x) \cos(4x)  }{ \tan(4x) }  \bigg]

We know that tan(x) is the ratio of sin(x) and cos(x). So, the limit becomes:

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\sin(4x) \cos(4x)  }{ \sin(4x) \div  \cos(4x)  }\bigg]

 \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\cos(4x)  }{1\div  \cos(4x)  }\bigg]

 \displaystyle \rm = \lim_{x \to0} \big[  2\cos^{2} (4x) \big]

Now substitute x = 0 to get the value of the limit:

 \displaystyle \rm = 2\cos^{2} (4 \times 0)

 \displaystyle \rm = 2\cos^{2} (0)

 \displaystyle \rm = 2 \times 1

 \displaystyle \rm = 2

Therefore:

 \displaystyle \rm \longrightarrow\lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) }  \bigg] = 2

★ Which is our required answer.

Learn More:

Standard limits.

\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0

\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1

\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1

\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1

\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0

\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1

\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1

\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1

\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1

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