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Prove that the function $f (x) = 5x-3$ is continuous at $x=0$, $x=-3$ and $x=5$.

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  • If $f$ is a real function on a subset of the real numbers and $c$ a point in the domain of $f$, then $f$ is continous at $c$ if $\lim\limits_{x\to c} f(x) = f(c)$.
Given $f(x) = 5x-3$.
$\textbf {Step 1}$
At $x=0, \; \lim\limits_{x\to 0} f(x) = \lim\limits_{x\to 0} 5x-3 = 5 \times 0 - 3 = -3$
$f(0) = 5 \times 0 - 3 = -3$
Since $\lim\limits_{x\to 0} f(x) = f(0)$, $f(x)$ is continous at $x=0$.
$\textbf {Step 2}$
At $x=-3, \; \lim\limits_{x\to -3} f(x) = \lim\limits_{x\to -3} 5x-3 = 5 \times -3 - 3 = -15-3 = -18$
$f(-3) = 5 \times -3 - 3 =-15-3=-18$
Since $\lim\limits_{x\to -3} f(x) = f(-3)$, $f(x)$ is continous at $x=-3$.
$\textbf {Step 3}$
At $x=-3, \; \lim\limits_{x\to 5} f(x) = \lim\limits_{x\to 5} 5x-3 = 5 \times 5 - 3 = 25-3 = 22$
$f(5) = 5 \times 5 - 3 =25-3 = 22$
Since $\lim\limits_{x\to 5} f(x) = f(5)$, $f(x)$ is continous at $x=5$.

 

answered Apr 4, 2013 by balaji.thirumalai
edited Apr 4, 2013 by balaji.thirumalai
 

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