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$\lim_{x\rightarrow 0}\frac{log(\sin 7x+\cos 7x)}{\sin 3x}\; equals:$

Option 1)
$\frac{1}{3}log7$

Option 2)
$\frac{7}{3}$

Option 3)
$\frac{14}{3}$

Option 4)
$\frac{1}{3}$

Answers (1)

As we learnt in

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow 0}\frac{log(sin7x+cos7x)}{sin3x}$

$\lim_{x\rightarrow 0}\frac{log\ cos7x(1+tan7x)}{sin3x}$

$\lim_{x\rightarrow 0}\frac{log\ cos7x}{sin3x}+\lim_{x\rightarrow 0}\frac{log(1+tan7x)}{sin3x}$

$\frac{-7sin7x}{cos7x\times 3cos3x}+\lim_{x\rightarrow 0}\frac{log(1+tan7x)}{tan7x}\times \frac{tan7x}{tan3x}$

$0+1\times \frac{7}{3}=\frac{7}{3}$

Option 1)

$\frac{1}{3}log7$

This option is incorrect

Option 2)

$\frac{7}{3}$

This option is correct

Option 3)

$\frac{14}{3}$

This option is incorrect

Option 4)

$\frac{1}{3}$

This option is incorrect

Posted by

Sabhrant Ambastha

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