Question

Find the value of k, for which f\left( x \right)=\left\\{ \begin{aligned} & \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\text{ if }-1\le x<0 \\\ & \dfrac{2x+1}{x-1}\text{ if 0}\le \text{x <1 } \\\ \end{aligned} \right\\} is continuous at x=0.

Answer 1

Apply the conditions of continuity on the function and then move with finding the right-hand limit and left-hand limit of the equation. Further, we will use the L’Hospital rule to find the definite value of the limit after that we will equate with the value obtained from putting the value of x as 0 in the function.

Complete step-by-step solution:
We will start with applying the condition of continuity to this function in order to find the value k;
Now we know that in general $f\left( x \right)$ is said to be continuous at x=a if it satisfies following conditions:-
Condition A: $f\left( x \right)$Should be defined at x=a it means $f\left( a \right)$should be a real number
Condition B. Limit of function should exist as x approaches a i.e. Right hand limit = Left hand limit 🡪 $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Condition C: Limit of functions as x approaches a should be equal to $f\left( a \right)$. $\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)=f\left( a \right)$
Now we have to find the value of k when function is continuous at x=0, since it is given in the question that $f\left( x \right)$ is continuous at x=0 therefore $f\left( x \right)$ will satisfy all the above conditions;
Applying condition one by one,
Condition A:
$f\left( x \right)$ should be defined at $x=a$ 🡪 $f\left( a \right)$should be a real number
Now we have $a=0$ with us, therefore $f\left( 0 \right)$ should be a real number:
For the given function: f\left( x \right)=\left\\{ \begin{aligned} & \dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\text{ if }-1\le x<0 \\\ & \dfrac{2x+1}{x-1}\text{ if 0}\le \text{x1 } \\\ \end{aligned} \right\\} ; we will take $\dfrac{2x+1}{x-1}\text{ }$ as this is defined at $x=0$ :
$f\left( x \right)=\dfrac{2x+1}{x-1}\text{ }$,
Now: $\to f\left( 0 \right)=\dfrac{2.0+1}{0-1}=\dfrac{1}{-1}=-1$

Condition B. Limit should exist as x approaches a : Right Hand Limit= Left Hand Limit
$\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$
Applying this condition to the given function, now here we will take up the first part of the function to find out the left hand limit that is $\dfrac{\sqrt{1+kx}-\sqrt{1-kx}}{x}\text{ }$as we are approaching ${{0}^{-}}$ , and this is defined at $-1\le x<0$
$\underset{h\to 0}{\mathop{\lim }}\,f\left( a-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+k(a-h)}-\sqrt{1-k\left( a-h \right)}}{a-h}$
Here we have a=0 , Therefore : $\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+k(0-h)}-\sqrt{1-k\left( 0-h \right)}}{0-h}\Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1-kh}-\sqrt{1+kh}}{0-h}=\dfrac{0}{0}$
Since the limit is of $\dfrac{0}{0}$ form we will use the L’Hospital Rule
L’Hospital Rule states that if $\phi (x)$ and $\psi (x)$takes the form$\dfrac{0}{0}$ then $\underset{x\to a}{\mathop{\lim }}\,\dfrac{\phi (x)}{\psi (x)}=\underset{x\to a}{\mathop{\lim }}\,\dfrac{\phi '(x)}{\psi '(x)}$
Applying the L’Hospital Rule into the limit below: $\begin{aligned} & \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+k(0+h)}-\sqrt{1-k\left( 0+h \right)}}{0+h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+kh}-\sqrt{1-kh}}{h} \\\ & \underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sqrt{1+kh}-\sqrt{1-kh}}{h}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{d\left( \sqrt{1+kh} \right)}{dh}-\dfrac{d\left( \sqrt{1-kh} \right)}{dh}}{\dfrac{dh}{dh}}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\dfrac{k}{2\sqrt{1+kh}}-\dfrac{\left( -k \right)}{2\sqrt{1-kh}}}{1} \\\ & =\dfrac{k}{2}+\dfrac{k}{2}\Rightarrow k \\\ \end{aligned}$
We will now find the right hand limit and for that we will take $\dfrac{2x+1}{x-1}\text{ }$, as we are approaching ${{0}^{+}}$ and for $x\ge 0$ we have $f\left( x \right)=\dfrac{2x+1}{x-1}\text{ }$
Now,

& \underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{2(a+h)+1}{\left( a+h \right)-1} \right) \\\ & \Rightarrow \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{2a+2h+1}{\left( a+h \right)-1} \right) \\\ \end{aligned}$$ Now, Here we have a=0 , Therefore : $\begin{aligned} & \underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{2a+2h+1}{a+h-1} \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \dfrac{2h+1}{h-1} \right)\Rightarrow \dfrac{+1}{-1}=-1 \\\ & \\\ \end{aligned}$ Condition C. Now according to this condition: $RHL=f\left( a \right)=LHL$ As given in the condition, equate the left hand limit to $f\left( a \right)$ and $RHL$ which both arre equal to -1 as found in Condition A and Condition B. Therefore, $RHL=f\left( a \right)=LHL\Rightarrow -1=-1=k$ **Hence the value of k is -1.** **Note:** Remember to check all the conditions of continuity before applying any further formula, even if one condition does not get satisfied then it will lead to a discontinuous function. Before applying the L’Hospital Rule ensure that f(x) is of the form $\dfrac{0}{0};\dfrac{\infty }{\infty }$ then continue differentiation till you get the definite limit. Also, keep in mind to differentiate the numerator and denominator separately.