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The function $f(x)=\left\{\begin{array}{cl}x+k, & x<3 \\ 4, & x=3 \\ 3 x-5, & x>3\end{array}\right.$ is continuous at $x=3$.

$\therefore f(3)=\lim _{x \rightarrow 3} f(x)$

$\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$

$\Rightarrow f(3)=\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3^{+}} f(x)$                  ...(1)

Now,

$f(3)=4$                         .....(2)

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}(x+k)=3+k$              ....(3)

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}(3 x-5)=3 \times 3-5=9-5=4$                     .....(4)

From (1), (2), (3) and (4), we have

$4=3+k=4$

$\Rightarrow 3+k=4$

$\Rightarrow k=4-3=1$

Thus, the value of k is 1.

If $f(x)=\left\{\begin{array}{cl}x+k, & x<3 \\ 4, & x=3 \\ 3 x-5, & x>3\end{array}\right.$ is continuous at $x=3$, then $k=$