Evaluate the following limits.
\[\underset{x\to 0}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}.\]
(a) \[\dfrac{1}{3}\]
(b) \[\dfrac{2}{3}\]
(c) \[\dfrac{5}{3}\]
(d) None of these

Hint: In this question, as both the right hand limit and the left hand limit exist and are equal we can say that the limit exists. Then by substituting the value of x directly in the given function we get the value of the limit.

Complete step-by-step answer:
LIMIT: Let y = f(x) be a function of x. If at x = a, f(x) takes indeterminate form, then we consider the value of the function which is very near to a. If these values tend to a definite unique number as x tends to a, then the unique number, so obtained is called the limit of f(x) at x = a and we write it as \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\].
Existence of limit: \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\] exists, if
\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\] and \[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\] both exist.
\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\]
Now, from the given question we have
\[\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}\]
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{3x+1}{x+3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}\]
Let us assume that the limit of the given function as L.
\[\Rightarrow L=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}\]
Now, by substituting the value of x in the given function we get,
\[\Rightarrow L=\dfrac{3\left( 0 \right)+1}{0+3}\]
Now, on further simplification we get,
\[\begin{align}
  & \Rightarrow L=\dfrac{0+1}{0+3} \\
 & \therefore L=\dfrac{1}{3} \\
\end{align}\]
Hence, the correct option is (a).

Note: It is important to note that the left hand limit and the right hand limit does not affect the limit value in this question because either when x approaches from left side or from the right side the value of the functions remains the same.
\[\Rightarrow \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{3x+1}{x+3}=\dfrac{1}{3}\]
We can directly get the value of the function which will be its limit without assuming any variable. It is important to note that in the numerator on multiplying 3 with 0 it becomes 0 and then we need to add 1 to that 0.