Lim x tends to 0 (sin 8x/tan 4x)
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Subject:
Math -
Author:
alysonconner -
Created:
1 year ago
Answers 1
Given Limit:
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) } \bigg][/tex]
If we substitute x = 0 directly, we get:
[tex] \displaystyle \rm =\dfrac{ \sin(0) }{ \tan(0) }[/tex]
[tex] \displaystyle \rm =\dfrac{0}{0}[/tex]
Which is an indeterminate form.
Consider again:
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) } \bigg][/tex]
We know that:
[tex] \bigstar \: \underline{ \boxed{ \rm \sin(2x) = 2 \sin(x) \cos(x) }}[/tex]
Using this result, we get:
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ \sin(2 \times 4x) }{ \tan(4x) } \bigg][/tex]
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\sin(4x) \cos(4x) }{ \tan(4x) } \bigg][/tex]
We know that tan(x) is the ratio of sin(x) and cos(x). So, the limit becomes:
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\sin(4x) \cos(4x) }{ \sin(4x) \div \cos(4x) }\bigg] [/tex]
[tex] \displaystyle \rm = \lim_{x \to0} \bigg[ \dfrac{ 2\cos(4x) }{1\div \cos(4x) }\bigg] [/tex]
[tex] \displaystyle \rm = \lim_{x \to0} \big[ 2\cos^{2} (4x) \big] [/tex]
Now substitute x = 0 to get the value of the limit:
[tex] \displaystyle \rm = 2\cos^{2} (4 \times 0)[/tex]
[tex] \displaystyle \rm = 2\cos^{2} (0)[/tex]
[tex] \displaystyle \rm = 2 \times 1[/tex]
[tex] \displaystyle \rm = 2[/tex]
Therefore:
[tex] \displaystyle \rm \longrightarrow\lim_{x \to0} \bigg[ \dfrac{ \sin(8x) }{ \tan(4x) } \bigg] = 2[/tex]
★ Which is our required answer.
Learn More:Standard limits.
[tex]\displaystyle\rm 1.\:\: \lim_{x\to0}\sin(x)=0[/tex]
[tex]\displaystyle\rm 2.\:\: \lim_{x\to0}\cos(x)=1[/tex]
[tex]\displaystyle\rm 3.\:\: \lim_{x\to0}\dfrac{\sin(x)}{x}=1[/tex]
[tex]\displaystyle\rm 4.\:\: \lim_{x\to0}\dfrac{\tan(x)}{x}=1[/tex]
[tex]\displaystyle\rm 5.\:\: \lim_{x\to0}\dfrac{1-\cos(x)}{x}=0[/tex]
[tex]\displaystyle\rm 6.\:\: \lim_{x\to0}\dfrac{\sin^{-1}(x)}{x}=1[/tex]
[tex]\displaystyle\rm 7.\:\: \lim_{x\to0}\dfrac{\tan^{-1}(x)}{x}=1[/tex]
[tex]\displaystyle\rm 8.\:\: \lim_{x\to0}\dfrac{\log(1+x)}{x}=1[/tex]
[tex]\displaystyle\rm 9.\:\: \lim_{x\to0}\dfrac{e^{x}-1}{x}=1[/tex]
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