Question

If $f\left( x \right)=\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3},x\ne 3$ is continuous at x=3, then which one of the following is correct?
a) f(3)=0
b) f(3)=1.5
c) f(3)=3
d) f(3)=-1.5

Answer 1

Hint:Simplify the numerator and denominator of $f(x)$ and cancel out the like terms. By using limits and derivatives, apply the limit $x\rightarrow3$ and find the value of $f(3)$.

Complete step-by-step answer:
Given, $f\left( x \right)=\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3}$
From the question it’s told that function is continuous at x=3 i.e. there will be no breakage of graph.

& \therefore \underset{x\to 3}{\mathop{\lim }}\,f\left( x \right)=f\left( 3 \right) \\\ & \Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-9}{{{x}^{2}}-2x-3}......\left( 1 \right) \\\ \end{aligned}$$ We know $$\left( {{x}^{2}}-9 \right)$$, can be written as $$\left( {{x}^{2}}-{{3}^{2}} \right)$$ where, $${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$$ $$\therefore {{x}^{2}}-{{3}^{2}}=\left( x-3 \right)\left( x+3 \right)......\left( 2 \right)$$ Similarly, $${{x}^{2}}+2x-3$$ can be simplified by finding the roots we can find the roots by using the quadratic equation, which is of the form $$a{{x}^{2}}+bx+c=0$$. $$\therefore $$Comparing $$\left( a{{x}^{2}}+bx+c \right)$$and $$\left( {{x}^{2}}+2x-3 \right)$$we can find that a=1, b=2 and c=-3. Substituting the values in the quadratic equation, $$\begin{aligned} & \dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\Rightarrow \dfrac{-2\pm \sqrt{{{2}^{2}}-4\times 1\times \left( -3 \right)}}{2\times 1} \\\ & =\dfrac{-2\pm \sqrt{4+12}}{2}=\dfrac{-2\pm \sqrt{16}}{2}=\dfrac{-2\pm 4}{2} \\\ \end{aligned}$$ $$\therefore $$We get two roots as $$\left( \dfrac{-2+4}{2} \right)$$and $$\left( \dfrac{-2-4}{2} \right)$$ $$\therefore $$Roots are (1, -3). $$\therefore $$$$\left( {{x}^{2}}+2x-3 \right)$$becomes $$\left( x-1 \right)\left( x+3 \right).......(3)$$ Substitute (2) and (3) in equation (1) $$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-9}{{{x}^{2}}+2x-3}=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\left( x+3 \right)\left( x-3 \right)}{\left( x-1 \right)\left( x+3 \right)}$$ Cancel out $(x+3)$ in the numerator and denominator = $$\underset{x\to 3}{\mathop{\lim }}\,\dfrac{x-3}{x-1}$$ Put x=3 $$=\dfrac{3-3}{3-1}=\dfrac{0}{2}=0$$ Any fraction with zero as numerator and any number as denominator is always equal to zero. $$\therefore f\left( 3 \right)=0$$ Hence, the correct option is (a) $$f\left( 3 \right)=0$$ Note: Find the roots of both numerator and denominator and then apply f(3) where x=3, to get the desired output.For these types of problems always try to factorize the given equation and simplify to cancel any common terms of both in numerator and denominator