Limit

lim⁡x→1exp⁡{cot⁡(πx)log⁡(sin⁡(πx)+1)}=lim⁡x→1exp⁡{cos⁡(πx)sin⁡(πx)log⁡(sin⁡(πx)+1)}=lim⁡x→1exp⁡{cos⁡(πx)}⏟direct sub⋅lim⁡x→1exp⁡{log⁡(sin⁡(πx)+1)sin⁡(πx)}⏟L’Hopital\lim_{x \to 1}\exp\left\{\cot(\pi x)\log(\sin(\pi x)+1)\right\}=\\ \lim_{x \to 1}\exp\left\{\frac{\cos(\pi x)}{\sin(\pi x)}\log(\sin(\pi x)+1)\right\}=\\ \underbrace{\lim_{x \to 1}\exp\{\cos(\pi x)\}}_{\text{direct sub}}\cdot \underbrace{ \lim_{x \to 1}\exp\left\{\frac{\log(\sin(\pi x)+1)}{\sin(\pi x)}\right\}}_{\text{L'Hopital}}
 
Let y=lim⁡x→1[1+sin⁡(πx)]cot⁡(πx)\displaystyle y = \lim_{x \to 1} [1+\sin(\pi x)]^{\cot(\pi x)}

ln⁡y=lim⁡x→1ln⁡([1+sin⁡(πx)]cot⁡(πx))\displaystyle \ln{y} = \lim_{x \to 1} \ln\bigg([1+\sin(\pi x)]^{\cot(\pi x)} \bigg)

ln⁡y=lim⁡x→1ln⁡[1+sin⁡(πx)]tan⁡(πx)\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{\ln[1+\sin(\pi x)]}{\tan(\pi x)}

now try applying L'Hopital
 
BigBeachBanana said:
lim⁡x→1exp⁡{cot⁡(πx)log⁡(sin⁡(πx)+1)}=lim⁡x→1exp⁡{cos⁡(πx)sin⁡(πx)log⁡(sin⁡(πx)+1)}=lim⁡x→1exp⁡{cos⁡(πx)}⏟direct sub⋅lim⁡x→1exp⁡{log⁡(sin⁡(πx)+1)sin⁡(πx)}⏟L’Hopital\lim_{x \to 1}\exp\left\{\cot(\pi x)\log(\sin(\pi x)+1)\right\}=\\ \lim_{x \to 1}\exp\left\{\frac{\cos(\pi x)}{\sin(\pi x)}\log(\sin(\pi x)+1)\right\}=\\ \underbrace{\lim_{x \to 1}\exp\{\cos(\pi x)\}}_{\text{direct sub}}\cdot \underbrace{ \lim_{x \to 1}\exp\left\{\frac{\log(\sin(\pi x)+1)}{\sin(\pi x)}\right\}}_{\text{L'Hopital}}
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I am confused about direct sub. How did you get that.
 
skeeter said:
Let y=lim⁡x→1[1+sin⁡(πx)]cot⁡(πx)\displaystyle y = \lim_{x \to 1} [1+\sin(\pi x)]^{\cot(\pi x)}

ln⁡y=lim⁡x→1ln⁡([1+sin⁡(πx)]cot⁡(πx))\displaystyle \ln{y} = \lim_{x \to 1} \ln\bigg([1+\sin(\pi x)]^{\cot(\pi x)} \bigg)

ln⁡y=lim⁡x→1ln⁡[1+sin⁡(πx)]tan⁡(πx)\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{\ln[1+\sin(\pi x)]}{\tan(\pi x)}

now try applying L'Hopital
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I don't understand how you got the third row.
 
Loki123 said:
I don't understand how you got the third row.
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ln⁡y=lim⁡x→1ln⁡[1+sin⁡(πx)]cot⁡(πx)\displaystyle \ln{y} = \lim_{x \to 1} \ln[1 + \sin(\pi x)]^{\cot(\pi x)}

ln⁡y=lim⁡x→1cot⁡(πx)⋅ln⁡[1+sin⁡(πx)]\displaystyle \ln{y} = \lim_{x \to 1} \cot(\pi x) \cdot \ln[1 + \sin(\pi x)]

ln⁡y=lim⁡x→11tan⁡(πx)⋅ln⁡[1+sin⁡(πx)]\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{1}{\tan(\pi x)} \cdot \ln[1 + \sin(\pi x)]

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skeeter said:
ln⁡y=lim⁡x→1ln⁡[1+sin⁡(πx)]cot⁡(πx)\displaystyle \ln{y} = \lim_{x \to 1} \ln[1 + \sin(\pi x)]^{\cot(\pi x)}

ln⁡y=lim⁡x→1cot⁡(πx)⋅ln⁡[1+sin⁡(πx)]\displaystyle \ln{y} = \lim_{x \to 1} \cot(\pi x) \cdot \ln[1 + \sin(\pi x)]

ln⁡y=lim⁡x→11tan⁡(πx)⋅ln⁡[1+sin⁡(πx)]\displaystyle \ln{y} = \lim_{x \to 1} \dfrac{1}{\tan(\pi x)} \cdot \ln[1 + \sin(\pi x)]

...
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How did you get lny?
 
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