Limit

[BigBeachBanana]

I took log with the base of 10 instead of ln

I've edited Skeeter's base e to base 10.
[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\log_{10}(1+\sin(\pi x))}{\tan(\pi x)}[/imath]

L’Hopital …

[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\frac{\pi \cos(\pi x)}{\log(10)(1+\sin(\pi x))}}{\pi \sec^2(\pi x)}[/imath]

[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\cos^3(\pi x)}{\log(10)(1+\sin(\pi x))}= \frac{-1}{\log(10)}[/imath]

[imath]y=10^{\frac{-1}{\log(10)}}=\frac{1}{e}[/imath]

 

[Loki123]

I've edited Skeeter's base e to base 10.
[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\log_{10}(1+\sin(\pi x))}{\tan(\pi x)}[/imath]

L’Hopital …

[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\frac{\pi \cos(\pi x)}{\log(10)(1+\sin(\pi x))}}{\pi \sec^2(\pi x)}[/imath]

[imath]\displaystyle \log_{10}{y} = \lim_{x \to 1} \dfrac{\cos^3(\pi x)}{\log(10)(1+\sin(\pi x))}= \frac{-1}{\log(10)}[/imath]

[imath]y=10^{\frac{-1}{\log(10)}}=\frac{1}{e}[/imath]

How did you get log after using L'Hopital. Doesn't it turn to xlna

 

[BigBeachBanana]

How did you get log after using L'Hopital. Doesn't it turn to xlna

They are. Notice I didn't write [imath]\log_{10}(10)[/imath], but simply [imath]\log(10)[/imath] which meant to be understood as [imath]\log_{e}(10)=\ln(10)[/imath]

 

[Loki123]

10log(10)−1=e1

Could you write this process a bit more detailed because I don't get it.

 

[BigBeachBanana]

Could you write this process a bit more detailed because I don't get it.

[math]10^{\frac{-1}{\ln 10}}=10^{\frac{-\ln e}{\ln 10}}=10^{-\log_{10}e}=e^{-1}[/math]Applying the change of base for log functions: [math]\log_{a}b=\frac{\log_{c}a}{\log_{c}b}[/math]