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Question Number 88896 by M±th+et£s last updated on 13/Apr/20

$$find$$$$\underset{x\to 0}{lim}\frac{ln(sin\left(3x\right)+cos\left(3x\right))}{ln(sin\left(x\right)+cos\left(x\right))}$$

Commented by abdomathmax last updated on 13/Apr/20

$$letf\left(x\right)=\frac{ln(sin\left(3x\right)+cos\left(3x\right))}{ln(sinx+cosx)}wehave$$$$sin\left(3x\right)+cos\left(3x\right)\sim 3x+1-\frac{{\left(3x\right)}^2}{2}=1+3x-\frac{9x^2}{2}$$$$andln(sin\left(3x\right)+cos\left(3x\right))\sim ln(1+3x-\frac{9x^2}{2})$$$$\sim 3x-\frac{9x^2}{2}$$$$sin\left(x\right)+cosx\sim x+1-\frac{x^2}{2}\Rightarrow$$$$ln(sinx+cosx)\sim ln(1+x-\frac{x^2}{2})\sim x-\frac{x^2}{2}\Rightarrow$$$$f\left(x\right)\sim \frac{3x-\frac{9x^2}{2}}{x-\frac{x^2}{2}}\Rightarrow f\left(x\right)\sim \frac{3-\frac{9x}{2}}{1-\frac{x}{2}}\Rightarrow$$$${lim}_{x\to 0}f\left(x\right)=3$$

Commented by M±th+et£s last updated on 13/Apr/20

$$nicesolutionthankyousir$$$$$$

Answered by TANMAY PANACEA. last updated on 13/Apr/20

$$\underset{x\to 0}{\lim }\frac{ln\{cos3x.(1+tan3x)\}}{ln\{cosx.(1+tanx)\}}$$$$=\underset{x\to 0}{\lim }\frac{lncos3x+ln(1+tan3x)}{lncosx+ln(1+tanx)}$$$$=\underset{x\to 0}{\lim }\frac{lncos3x+\frac{ln(1+tan3x)}{tan3x}\times \frac{tan3x}{3x}\times 3x}{lncosx+\frac{ln(1+tanx)}{tanx}\times \frac{tanx}{x}\times x}$$$$nowwhenx\to 0$$$$lncos3x\to 0lncosx\to 0$$$$\underset{x\to 0}{\lim }\frac{0+1\times 1\times 3x}{0+1\times 1\times x}=3$$$$$$$$let$$$$\underset{x\to 0}{\lim }$$

Commented by M±th+et£s last updated on 13/Apr/20

$$niceworkthankyoiverymuchsir$$

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