How to discuss the continuity of this function because it doesn't give a range in the first & third part.
The function is $F\colon [0,1] \to \mathbf R$ with
$$F(x)= \begin{cases} \cos x, \quad & x=0 \\ \frac{x \ln x}{x-1}, & 0
How to discuss the continuity of a function?
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$\begingroup$
limits
continuity
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2What is your function? – 2017-02-07
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0F(x), it should be cosx when x = 0, it should be (x/1-x)ln x when 0
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0Okay, here we go. So the function is $F\colon [0,1] \to \mathbf R$ with $$F(x)= \begin{cases} \cos x, \quad & x=0 \\ \frac{x \ln x}{x-1}, & 0
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0It's not clear. It shows me something like a programming code – 2017-02-07
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0Maybe refresh the site. I dont know. Have a look at my answer please. This might help. – 2017-02-07
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0@NiklasHebestreit can you give it as a separate answer – 2017-02-07
1 Answers
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I assume we are analyzing the function $F\colon [0,1] \to \mathbf R$ with
$$F(x)= \begin{cases} \cos x, \quad & x=0 \\ \frac{x \ln x}{x-1}, & 0 Hint: To check where $F$ is continuous you only have to check the critical points $0$ and $1$. Therefore you have to check if
$$F(0)=\cos 0 =1 \stackrel{?}{=} \lim_{x\to 0+ \\ 0
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0I don't know how to calculate the value of f(x) when 0
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0There is nothing to calculate. The value is given. – 2017-02-07
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0Where....? I don't understand – 2017-02-07
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0Can I please know that this is continuous or not? – 2017-02-07
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0Which part of my answer dont you understand and why? I really want to help you! – 2017-02-07
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00
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0Please read my answer **carefully** and think about it. – 2017-02-07
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0But still I can't understand – 2017-02-07
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0I got the answer... yes... it is continuous in that range – 2017-02-07