Find the value of k that makes f(x) continuous at x = 3
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Given:
The function is,
[tex]f(x)=f(x)=\begin{cases}\frac{x-3}{x^2+2x-15},x\ne3 \\ k,x=3\end{cases}[/tex]As the given function is continous at x= 3 ,
[tex]\begin{gathered} \lim _{x\to3}f(x)=k \\ \lim _{x\to3}(\frac{x-3}{x^2+2x-15})=k \end{gathered}[/tex]