Question

Define \[f:\left[ -\dfrac{1}{2},\infty \right)\to R\]by \[f\left( x \right)=\sqrt{1+2x},x\in \left[ -\dfrac{1}{2},\infty \right),\]then compute \[\underset{x\to -{{\dfrac{1}{2}}^{+}}}{\mathop{\lim }}\,f\left( x \right)\]. Hence find \[\underset{x\to -\dfrac{1}{2}}{\mathop{\lim }}\,f\left( x \right)\].

Answer 1

Hint: In order to define the function, we need to find the points of discontinuity of the function, this can be found by putting \[f\left( x \right)=\sqrt{1+2x}\] equal to zero. Then, finding limits of \[f\left( x \right)\] at values of \[x\] slightly greater than \[-\dfrac{1}{2}\] and that at \[-\dfrac{1}{2}\].

Formula used:
In order to draw the graph, the various points are considered and plotted. The domain for the function is \[\left[ -\dfrac{1}{2},\infty \right),\left\{ x\left| x \right.\ge -\dfrac{1}{2} \right\}\] as the interval over which function is defined is given as \[f:\left[ -\dfrac{1}{2},\infty \right)\to R\] in the question.
To find the, \[\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\]
\[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)\] and \[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)\] should exist and coincide.

Complete step by step solution:
Equating \[f\left( x \right)\] to zero will give the roots
\[\sqrt{1+2x}=0\]
Squaring both sides
\[\begin{align}
  & 1+2x=0 \\
 & x=-\dfrac{1}{2} \\
\end{align}\]
So, the function is discontinuous at the point \[x=-\dfrac{1}{2}\]
If we plot a graph using these points,

x	y
-0.5	0
0	1
1	1.73
2	2.24

The domain of the function is found by finding where the equation is defined. The range is the set of values that correspond to the domain.
So, the domain is \[\left[ -\dfrac{1}{2},\infty \right),\left\{ x\left| x \right.\ge -\dfrac{1}{2} \right\}\]
And the range is: \[\left[ 0,\infty \right),\left\{ y\left| y \right.\ge 0 \right\}\]
Now, in order to find \[\underset{x\to -{{\dfrac{1}{2}}^{+}}}{\mathop{\lim }}\,f\left( x \right)\]
We solve it in this manner,
\[x\leftrightarrow -\dfrac{1}{2}+h\]
Where \[h\to 0\]
\[\begin{align}
  & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,f\left( -\dfrac{1}{2}+h \right) \\
 & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\sqrt{1+2\left( -\dfrac{1}{2}+h \right)} \\
 & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\sqrt{1-1+h} \\
 & \Rightarrow 0 \\
\end{align}\]
Now, the range is from\[\left[ -\dfrac{1}{2},\infty \right)\], now as the left hand limit does not lie in the range. The left hand limit does not exist.
Now for \[\underset{x\to -\dfrac{1}{2}}{\mathop{\lim }}\,f\left( x \right)\] to exist, the left hand limit and the right hand limit should coincide. However, it is not so, in this case. So, the limit \[\underset{x\to -\dfrac{1}{2}}{\mathop{\lim }}\,f\left( x \right)\] does not exist.

Note: The function is discontinuous at \[x=-\dfrac{1}{2}\] means that the curve is not continuous at this point or makes a jump or is broken .If the interval contained the values less than \[-\dfrac{1}{2}\], then the left hand limit would have existed and also the \[\underset{x\to -\dfrac{1}{2}}{\mathop{\lim }}\,f\left( x \right)\].