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    Please, please please help me with this questions. Really struggling as i have been off uni ill for over a week now and missed alot of classes to make it worse this assesment is due in asap!!

    Many thanks, Hanna. x
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    For the first one

    \displaystyle\lim_{x \rightarrow 1^-} f(x) = 1

    \displaystyle\lim_{x \rightarrow 1^+} f(x) = 1

    So you can conclude the first function is continuous at 1. The way to view the above, is you're interested in knowing if the two functions 'meet' at x=1. So when x \rightarrow 1^- we're looking at the values of x tending to 1, but from the left (so using the function for when x<1) and then doing it for the values to the right but obviously using the function for when x \geq 1

    (Original post by K80dol)
    Since you've missed a few lectures I'd just like to add something to what Noble has said: continuity at a does not only rely on both sides of the curve meeting at a, the function also needs to be defined at a and if so, its value there needs to agree with both limits. Here is an added example:

    f(x)=\dfrac{\sin (x-1)}{x-1}

    \displaystyle\lim_{x\to 1^-}f(x)=\lim_{x\to 1^+}f(x)=1 yet f is discontinuous at 1 because of the obvious issue with f(1)

    If we defined f(1), like what is done in the third exercise of yours, then continuity will rely on what value we choose:

    f(x) = \left\{ \begin{array}{l l}\dfrac{\sin (x-1)}{x-1} & x\neq 1\\1 & x=1\end{array} \right.\quad\Rightarrow f\;\text{is continuous at}\;1

    On the other hand:

    f(x) = \left\{ \begin{array}{l l}\dfrac{\sin (x-1)}{x-1} & x\neq 1\\0 & x=1\end{array} \right.\quad\Rightarrow f\;\text{is discontinuous at}\;1
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