Question

Let \[g\left( x \right)=\left\{ \begin{align}
  & \dfrac{\sqrt{x+1}-1}{\sqrt{x}},x>0 \\
 & 0,x=0 \\
\end{align} \right.\]
Then g is
(a) Continuous and differentiable at \[x=0\]
(b) Continuous and differentiable for all \[x>0\]
(c) Continuous for all \[x>0\]
(d) Not right differentiable at \[x=0\]

Answer 1

Hint: Check continuity and differentiability for the given function around 0 by applying the limits.

Complete step-by-step answer:
We have the function of the form \[g\left( x \right)=\dfrac{\sqrt{x+1}-1}{\sqrt{x}}\] for \[x>0\] and \[g\left( x \right)=0\] at \[x=0\].
We will begin by checking the continuity of the function around 0.
To check the continuity of the function \[g\left( x \right)\] around point a, we have \[\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,g\left( a+h \right)=g\left( a \right)\] and \[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,g\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,g\left( a-h \right)=g\left( a \right)\].
We will find the continuity around point \[a=0\].
As \[g\left( x \right)\] is defined only for \[x\ge 0\], we need to check only the limit on the right side.
Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,g\left( 0+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{0+h+1}-1}{\sqrt{0+h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{\sqrt{h}}.....\left( 1 \right)\].
As \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{\sqrt{h}}=\dfrac{0}{0}\] , we will use L’Hopital Rule to find the limit which states that if \[\underset{h\to a}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{v\left( h \right)}=\dfrac{0}{0}\] then, we solve the limit by \[\underset{h\to a}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{v\left( h \right)}=\underset{h\to a}{\mathop{\lim }}\,\dfrac{u'\left( h \right)}{v'\left( h \right)}=\dfrac{u'\left( a \right)}{v'\left( a \right)}\].
Substituting \[u\left( h \right)=\sqrt{h+1}-1,v\left( h \right)=\sqrt{h}\] and \[\underset{h\to 0}{\mathop{\lim }}\,\] we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{v\left( h \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u'\left( h \right)}{v'\left( h \right)}\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{\sqrt{h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( \sqrt{h+1}-1 \right)}{\dfrac{d}{dh}\left( \sqrt{h} \right)}.....\left( 2 \right)\].
Thus, we will differentiate the functions \[u\left( h \right)\] and \[v\left( h \right)\].
To find the derivative of the function \[y={{\left( x+a \right)}^{n}}+b\], we have \[\dfrac{dy}{dx}=n{{\left( x+a \right)}^{n-1}}\].
Thus, we have \[\dfrac{d}{dh}\left( \sqrt{h+1}-1 \right)=\dfrac{1}{2\sqrt{h+1}}.....\left( 3 \right)\] by substituting \[a=1,n=\dfrac{1}{2},b=-1\] in the above equation.
Similarly, we have \[\dfrac{d}{dh}\left( \sqrt{h} \right)=\dfrac{1}{2\sqrt{h}}.....\left( 4 \right)\] by substituting \[a=0,n=\dfrac{1}{2},b=0\] in the above equation.
Substituting equation \[\left( 3 \right)\] and \[\left( 4 \right)\] in equation \[\left( 2 \right)\], we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{\sqrt{h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( \sqrt{h+1}-1 \right)}{\dfrac{d}{dh}\left( \sqrt{h} \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{2\sqrt{1+h}}}{\dfrac{1}{2\sqrt{h}}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h}}{\sqrt{h+1}}.....\left( 5 \right)\].
Now, applying the limit, we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h}}{\sqrt{h+1}}=0.....\left( 6 \right)\].
Using equation \[\left( 1 \right)\],\[\left( 2 \right)\], \[\left( 5 \right)\] and \[\left( 6 \right)\], we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{\sqrt{h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h}}{\sqrt{h+1}}=0\].
Thus, we have \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,g\left( x \right)=0=g\left( 0 \right)\].
Hence, the function \[g\left( x \right)\] is continuous for \[x\ge 0\].
Now, we will check the differentiability of \[g\left( x \right)\].
For \[x>0\], we have \[g\left( x \right)=\dfrac{\sqrt{x+1}-1}{\sqrt{x}}\]. We observe that \[g\left( x \right)\] is a composition of polynomials which is differentiable everywhere except at points where the function tends to \[\infty \] which happens when the denominator tends to 0.
But for \[x>0\], the denominator of \[g\left( x \right)\] never equals zero.
Hence, \[g\left( x \right)\] is continuous and differentiable for \[x>0\].
We will now check the differentiability of \[g\left( x \right)\] at \[x=0\].
To check the differentiability of the function \[g\left( x \right)\] around point a, we have \[g'\left( a \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( a+h \right)-g\left( a \right)}{h}\] and \[g'\left( a \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( a-h \right)-g\left( a \right)}{-h}\].
We will find the continuity around point \[a=0\].
As \[g\left( x \right)\] is defined only for \[x\ge 0\], we need to check only the limit on the right side.
We know that \[g\left( x \right)=0\] at \[x=0\].
Thus, we have \[\dfrac{d}{dx}g\left( x \right){{|}_{x=0}}=g'\left( 0 \right) \left( 7 \right)\].
Now, we have \[g'\left( a \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( a+h \right)-g\left( a \right)}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{\sqrt{h+1}-1}{\sqrt{h}}-0}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{h\sqrt{h}}.....\left( 8 \right)\].
As \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{h\sqrt{h}}=\dfrac{0}{0}\] , we will use L’Hopital Rule to find the limit which states that if \[\underset{h\to a}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{w\left( h \right)}=\dfrac{0}{0}\] then, we solve the limit by \[\underset{h\to a}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{w\left( h \right)}=\underset{h\to a}{\mathop{\lim }}\,\dfrac{u'\left( h \right)}{w'\left( h \right)}=\dfrac{u'\left( a \right)}{w'\left( a \right)}\].
Substituting \[u\left( h \right)=\sqrt{h+1}-1,w\left( h \right)=h\sqrt{h}={{h}^{\dfrac{3}{2}}}\] and\[\underset{h\to 0}{\mathop{\lim }}\,\]we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u\left( h \right)}{w\left( h \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{u'\left( h \right)}{w'\left( h \right)}\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{h\sqrt{h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( \sqrt{h+1}-1 \right)}{\dfrac{d}{dh}\left( {{h}^{\dfrac{3}{2}}} \right)}.....\left( 9 \right)\].
To find the derivative of the function \[y={{\left( x+a \right)}^{n}}+b\], we have \[\dfrac{dy}{dx}=n{{\left( x+a \right)}^{n-1}}\].
We have \[\dfrac{d}{dh}\left( {{h}^{\dfrac{3}{2}}} \right)=\dfrac{3\sqrt{h}}{2}.....\left( 10 \right)\] by substituting \[a=0,n=\dfrac{3}{2},b=0\] in the above equation.
Substituting equation \[\left( 3 \right)\] and \[\left( 10 \right)\] in equation \[\left( 9 \right)\], we have \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{h+1}-1}{h\sqrt{h}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d}{dh}\left( \sqrt{h+1}-1 \right)}{\dfrac{d}{dh}\left( {{h}^{\dfrac{3}{2}}} \right)}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{1}{2\sqrt{1+h}}}{\dfrac{3\sqrt{h}}{2}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{3\sqrt{h}\sqrt{h+1}}\].
We observe that \[\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{3\sqrt{h}\sqrt{h+1}}=\dfrac{1}{0}=\infty .....\left( 11 \right)\].
Hence, using equation \[\left( 7 \right)\],\[\left( 8 \right)\], \[\left( 9 \right)\] and \[\left( 11 \right)\], we have \[g'\left( a \right)=0\ne \underset{h\to 0}{\mathop{\lim }}\,\dfrac{g\left( a+h \right)-g\left( a \right)}{h}=\infty \].
Hence, \[g\left( x \right)\] is continuous but not differentiable for \[x\ge 0\], which is option (c), (d).

Note: It’s necessary to use the L'Hopital Rule to find the limits. If we will simply apply the limits, we will get an incorrect answer. Also, one must check the condition of continuity and differentiability using First Principle.