Limit

[Loki123]

How do I find the limit? CtgPi is giving me problems.

 

[BigBeachBanana]

[math]\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}}[/math]

 

[skeeter]

Let [imath]\displaystyle y = \lim_{x \to 1} [1+\sin(\pi x)]^{\cot(\pi x)}[/imath]

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

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

now try applying L'Hopital

 

[Loki123]

[math]\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}}[/math]

I am confused about direct sub. How did you get that.

 

[BigBeachBanana]

Its

I am confused about direct sub. How did you get that

The notes are for you what to do, not how I got there.

 

[Loki123]

Let [imath]\displaystyle y = \lim_{x \to 1} [1+\sin(\pi x)]^{\cot(\pi x)}[/imath]

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

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

now try applying L'Hopital

I don't understand how you got the third row.

 

[Loki123]

Its

The notes are for you what to do, not how I got there.

I don't understand them.

 

[BigBeachBanana]

I don't understand them.

[math]\lim g(x)h(x)=\lim g(x)\times \lim h(x)[/math]

 

[Loki123]

[math]\lim g(x)h(x)=\lim g(x)\times \lim h(x)[/math]

What do I do with that?

 

[BigBeachBanana]

What do I do with that?

You asked my how I got the direct substitution part in post #2. That’s the property I applied. Look at post 2 again.

 

[Loki123]

You asked my how I got the direct substitution part in post #2. That’s the property I applied. Look at post 2 again.

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

 

[skeeter]

I don't understand how you got the third row.

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

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

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

...

 

[Loki123]

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

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

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

...

How did you get lny?

 

[Deleted member 4993]

How did you get lny?

Look at response #3

 

[Loki123]

Look at response #3

Can you just "multiply" both sides with ln even with limits involved?

 

[skeeter]

Can you just "multiply" both sides with ln even with limits involved?

it’s not multiplying … you’re taking the natural log of both sides

 

[Loki123]

it’s not multiplying … you’re taking the natural log of both sides

What does natural log mean?

 

[Loki123]

What does natural log mean?

And why does it work while my way does not

 

[BigBeachBanana]

And why does it work while my way does not

What you’re doing is not different than what I’m doing. You need to do an extra step of algebraic manipulation.

 

[Loki123]

What you’re doing is not different than what I’m doing. You need to do an extra step of algebraic manipulation.

Why am I getting a wrong answer? Your way is at the top, mine is at the bottom.