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# (Solved): Consider the function $f(x)=\left\{\begin{array}{ll} 7 \cdot x+4 & x-\frac{1}{7} \end{array}\righ ... 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 $$x=-\frac{1}{7}$$.

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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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