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

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

& 1+2x=0 \\\ & x=-\dfrac{1}{2} \\\ \end{aligned}$$ So, the function is discontinuous at the point $$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{aligned} & \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{aligned}$$ Now, the range is from$$\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)$$.