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Consider the function \[ f(x)=\left\{\begin{array}{ll} 7 \cdot x+4 & x<-\frac{1}{7} \\ 4 & x=-\frac{1}{7} \\ 147 \cdot x^{2} & x>-\frac{1}{7} \end{array}\right. \] Tick all of the following statements that are correct. \[ \lim _{x \rightarrow\left(-\frac{1}{7}\right)^{+}} f(x)=4 \] \( \lim _{x \rightarrow-\frac{1}{7}} f(x) \) exists. \( f \) has a jump discontinuity at \( x=-\frac{1}{7} \). \( \lim _{x \rightarrow\left(-\frac{1}{7}\right)} f(x)=3 \) \( f \) is continuous at \( x=-\frac{1}{7} \). \[ \lim _{x \rightarrow-\frac{1}{7}} f(x)=f\left(-\frac{1}{7}\right) \]
\( f \) has a removable discontinuity at \( x=-\frac{1}{7} \). \( f \) is discontinuous at \( x=-\frac{1}{7} \).

Solution: The right hand limit of f(x) at x=?17 is limx??(17)+f(x)=limh?0f(?17+h) =limh?0147×(?17+h)2---------(1) =147×(?17+0)2 =147×149 =3.. Similarl

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