\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

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