Now f is defined at x = 0
and f(0) = 5(0) - 3 = 0 - 3 = -3
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(0) = -3\\ \ \ \ \text{x}\rightarrow0$
$\therefore$ f is continous at x = 0
Now f is defined at x = -3
and f(-3) = 5(-3) - 3 = -15 - 3 = -18
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(-3) = -18\\ \ \ \ \text{x}\rightarrow-3$
$\therefore$ f is continous at x = -3
Now f is defined at x = 5
and f(5) = 5(5) - 3 = 25 - 3 = 22
$\therefore \ \ \text{Lt}\ \ \ \ \ \ \ \ \ \ \ \text{f(x)} = \text{f}(5) = 22\\ \ \ \ \text{x}\rightarrow5$
$\therefore$ f is continous at x = 5
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$x \cos \text{y dy} = ( xe^{x} \log x + e^{x}) dx$