Limit

Loki123 said:
Idk what you see as a, n and x so the order might be different
But
x=e^cotpix
a=e
n=cotpix
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I mean from here:
[math]\ln e^{\cot(\pi x)}[/math]Rewrite:
[math]\ln e^{\cot(\pi x)}=\log_{\red{e}^\orange{1}}\purple{e^{\cot(\pi x)}}[/math]so
[math]"x"=e^{\cot(\pi x)}\\ "a"=e\\ "n"=1\\[/math]Apply the property you presented, you get:
[math]\frac{1}{1} \log_{e}e^{\cot(\pi x)}=\ln e^{\cot(\pi x)}[/math]which is what you started with, so this property doesn't help you and you applied it incorrectly.
 
BigBeachBanana said:
I mean from here:
[math]\ln e^{\cot(\pi x)}[/math]Rewrite:
[math]\ln e^{\cot(\pi x)}=\log_{\red{e}^\orange{1}}\purple{e^{\cot(\pi x)}}[/math]so
[math]"x"=e^{\cot(\pi x)}\\ "a"=e\\ "n"=1\\[/math]Apply the property you presented, you get:
[
which is what you started with, so this property doesn't help you.
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That's not what I did. Look
IMG_20220417_223317.jpg
 
Loki123 said:
That's not what I did. Look
View attachment 32218
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Now, I understand. Your mistake is here.

Screen Shot 2022-04-17 at 3.39.57 PM.png

Apply the property you should get:
[imath]\log_{e^{\tan(\pi x)}}(1+\sin (\pi x))[/imath] NOT [imath]\ln_{e^{\tan(\pi x)}}(1+\sin (\pi x))[/imath].
Apply the limit you get [math]e^{\log_{1^1}(1+\sin(\pi))}=e^{\log_{1}(1+\sin(\pi))}[/math]What's the value of [imath]\log_{1}(1+\sin(\pi))?[/imath]
 
BigBeachBanana said:
Now, I understand. Your mistake is here.

View attachment 32219

Apply the property you should get:
[imath]\log_{e^{\tan(\pi x)}}(1+\sin (\pi x))[/imath] NOT [imath]\ln_{e^{\tan(\pi x)}}(1+\sin (\pi x))[/imath].
Apply the limit you get [math]e^{\log_{1^1}(1+\sin(\pi))}=e^{\log_{1}(1+\sin(\pi))}[/math]What's the value of [imath]\log_{1}(1+\sin(\pi))?[/imath]
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It can be 1 or 0
 
Loki123 said:
It can be 1 or 0
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Not quite.
[imath]\log_{1}(1+\sin(\pi))=\log_{1}1=\text{undefined}[/imath]. Why?
Let [imath]y=\log_1(1)\implies 1^y=1[/imath]
For what value of y is true is the equation true? The answer is any value of y would make this equation true, and this is a bad thing for a function so we leave it undefined.
 
BigBeachBanana said:
Not quite.
[imath]\log_{1}(1+\sin(\pi))=\log_{1}1=\text{undefined}[/imath]. Why?
Let [imath]y=\log_1(1)\implies 1^y=1[/imath]
For what value of y is true? The answer is any value of y would make this equation true, this is a bad thing for a function so we leave it undefined.
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So I cant get a good answer my way??
 
Loki123 said:
So I cant get a good answer my way??
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I don't see how, maybe @skeeter or someone else can.
When you evaluate the limit and get something undefined, you gotta try a different approach/method like L'Hopital to make it "defined". I haven't seen anyone use your property before, quite interesting though.
 
BigBeachBanana said:
I don't see how, maybe @skeeter or someone else can.
When you evaluate the limit and get something undefined, you gotta try a different approach/method like L'Hopital to make it "defined". I haven't seen anyone use your property before, quite interesting though.
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What about this
I also get just eIMG_20220417_233626.jpg
 
Loki123 said:
Last part is wrong but how do I get e^-1
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the natural log, [imath]\ln[/imath], is base e …

[imath]\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{\ln(1+\sin(\pi x)}{\tan(\pi x)}[/imath]

L’Hopital …

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

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

[imath]\ln{y} = -1 \implies y = e^{-1} = \dfrac{1}{e}[/imath]
 
skeeter said:
the natural log, [imath]\ln[/imath], is base e …

[imath]\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{\ln(1+\sin(\pi x)}{\tan(\pi x)}[/imath]

L’Hopital …

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

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

[imath]\ln{y} = -1 \implies y = e^{-1} = \dfrac{1}{e}[/imath]
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I did that. What about #33 and #34
 
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