Determining if a function is continuous
\begin{array}{l} f(x)=\begin{cases} \dfrac{|x|}x & \text{if $x\ne 0$}\\ 0 & \text{if $x = 0$} \end{cases}\\ g(x)=\begin{cases} \dfrac{\sin x} x & \text{if $x\ne 0$}\\ 1 & \text{if $x = 0$} \end{cases}\\ h(x)=\begin{cases} x^2+2x & \text{if $x\ge 1$}\\ 1+2\cos(x-1) & \text{if $x < 1$} \end{cases} \end{array}
I know that a function is continuous at $a$ iff $$\lim_{x \to a} f(x) = f(a).$$
But I can't really figure out when a function is continuous everywhere.
Can you tell me the way I should approach this, without plotting a graph?
For the first one, I know that the form of the plot of the function looks like this:
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$\begingroup$Continuity of a function is defined if it is continuous in the entire domain , such that for every $a$ , $f(a) = \lim_{x \rightarrow a}f(x) $ should exist . Now for $g(x)$ you can verify that the function will be continuous at every point for $a \ne 0$ ie you can verify that if $a\ne0$ then $\lim_{x\rightarrow a} \frac{sin(x)}{x} = \frac{sin(a)}{a}$ which is equal to $f(a) = \frac{sin(a)}{a}$ . But the only point where one can be suspicious about the function being discontinous is at the point $a=0$ because there the denominator will become $0$ . So we will evaluate the limit as $x \rightarrow 0$ which for this case is equal to $1$ and the value of $f(x)$ at this point ie $f(0) = 1$ is given to be $1$ , hence the function is continuous everywhere .
$\endgroup$ $\begingroup$The first one is not continuous, because $\lim_{x\rightarrow0^+} f(x) = 1$ and $\lim_{x\rightarrow0^-} = -1$ but $f(0) = 0$.
The second one is continuous. The only limit point you need to check is $x=0$ because $\sin(x) / x$ is obviously continuous everywhere else. And yes, $\lim_{x\rightarrow0} \sin(x)/x = 1$.
The third one is also continuous. Again, the only point you have to check is the junction at $x=1$. $f(1) = \lim_{x\rightarrow1^+} f(x) = 3$ because $x^2 + 2x$ is continuous. So you need to check $\lim_{x\rightarrow1^-} 1 + 2\cos(x-1) = 1 + 2\cos(1-1) = 3$.
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