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Question: If \[f\left( 9 \right) = 9\] and \[f'\left( 9 \right) = 1\], then \[\mathop {\lim }\limits_{x \to 9}...

If f(9)=9f\left( 9 \right) = 9 and f′(9)=1f'\left( 9 \right) = 1, then lim⁡x→93−f(x)3−x\mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} is equal to
(a) 0
(b) 1
(c) −1 - 1
(d) None of these

Explanation

Solution

Here, we need to find the value for the given expression. Using the given information, algebraic identities, rationalising, and the formula of a derivative of a function at a point, we will simplify the given expression and find the required value.

Formula used:
We will use the following formulas:
1.The product of the sum and difference of two numbers is given by the algebraic identity (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}.
2.The derivative of a function f(x)f\left( x \right) at x=ax = a is given by the formula f′(a)=lim⁡h→0f(a+h)−f(a)hf'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}. This can be written as f′(a)=lim⁡x→af(x)−f(a)x−af'\left( a \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}, where h=x−ah = x - a.

Complete step-by-step answer:
We will simplify the given expression with the limit to find its value.
Factoring out −1 - 1 from the numerator and denominator of the expression lim⁡x→93−f(x)3−x\mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }}, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9−1[f(x)−3]−1[x−3]\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{ - 1\left[ {\sqrt {f\left( x \right)} - 3} \right]}}{{ - 1\left[ {\sqrt x - 3} \right]}}
Simplifying the expression, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−3x−3\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt {f\left( x \right)} - 3}}{{\sqrt x - 3}}
Multiplying and dividing the expression by x+3\sqrt x + 3, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−3x−3×(x+3x+3) ⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−3(x−3)(x+3)×(x+3)\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt {f\left( x \right)} - 3}}{{\sqrt x - 3}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt x + 3}}} \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt {f\left( x \right)} - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} \times \left( {\sqrt x + 3} \right)\end{array}
Multiplying and dividing the expression by f(x)+3\sqrt {f\left( x \right)} + 3, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−3(x−3)(x+3)×(x+3)×(f(x)+3f(x)+3) ⇒lim⁡x→93−f(x)3−x=lim⁡x→9(f(x)−3)(f(x)+3)(x−3)(x+3)×(x+3)×(1f(x)+3)\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\sqrt {f\left( x \right)} - 3}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} \times \left( {\sqrt x + 3} \right) \times \left( {\dfrac{{\sqrt {f\left( x \right)} + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} \times \left( {\sqrt x + 3} \right) \times \left( {\dfrac{1}{{\sqrt {f\left( x \right)} + 3}}} \right)\end{array}
Simplifying the expression, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9(f(x)−3)(f(x)+3)(x−3)(x+3)×(x+3f(x)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)
Substituting a=xa = \sqrt x and b=3b = 3 in the identity (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}, we get
⇒(x−3)(x+3)=(x)2−32\Rightarrow \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) = {\left( {\sqrt x } \right)^2} - {3^2}
Applying the exponents on the terms, we get
⇒(x−3)(x+3)=x−9\Rightarrow \left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) = x - 9
Substituting a=f(x)a = \sqrt {f\left( x \right)} and b=3b = 3 in the identity (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}, we get
⇒(f(x)−3)(f(x)+3)=(f(x))2−32\Rightarrow \left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right) = {\left( {\sqrt {f\left( x \right)} } \right)^2} - {3^2}
Applying the exponents on the terms, we get
⇒(f(x)−3)(f(x)+3)=f(x)−9\Rightarrow \left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right) = f\left( x \right) - 9
Substituting (x−3)(x+3)=x−9\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) = x - 9 and (f(x)−3)(f(x)+3)=f(x)−9\left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right) = f\left( x \right) - 9 in the equation lim⁡x→93−f(x)3−x=lim⁡x→9(f(x)−3)(f(x)+3)(x−3)(x+3)×(x+3f(x)+3)\mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{\left( {\sqrt {f\left( x \right)} - 3} \right)\left( {\sqrt {f\left( x \right)} + 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right), we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−9x−9×(x+3f(x)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - 9}}{{x - 9}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)
It is given that f(9)=9f\left( 9 \right) = 9.
Substituting 9=f(9)9 = f\left( 9 \right) in the equation, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−f(9)x−9×(x+3f(x)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)
Rewriting the expression, we get
⇒lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−f(9)x−9×lim⁡x→9(x+3f(x)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} \times \mathop {\lim }\limits_{x \to 9} \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)
The derivative of a function f(x)f\left( x \right) at x=ax = a is given by the formula f′(a)=lim⁡h→0f(a+h)−f(a)hf'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}.
This can be written as f′(a)=lim⁡x→af(x)−f(a)x−af'\left( a \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}, where h=x−ah = x - a.
Therefore, the derivative of the function f(x)f\left( x \right) at x=9x = 9 can be written as f′(9)=lim⁡x→9f(x)−f(9)x−9f'\left( 9 \right) = \mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}}.
Substituting lim⁡x→9f(x)−f(9)x−9=f′(9)\mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} = f'\left( 9 \right) in the equation lim⁡x→93−f(x)3−x=lim⁡x→9f(x)−f(9)x−9×lim⁡x→9(x+3f(x)+3)\mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = \mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} \times \mathop {\lim }\limits_{x \to 9} \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right), we get
⇒lim⁡x→93−f(x)3−x=f′(9)×lim⁡x→9(x+3f(x)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = f'\left( 9 \right) \times \mathop {\lim }\limits_{x \to 9} \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right)
Applying the limit x=9x = 9, we get
⇒lim⁡x→93−f(x)3−x=f′(9)×(9+3f(9)+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = f'\left( 9 \right) \times \left( {\dfrac{{\sqrt 9 + 3}}{{\sqrt {f\left( 9 \right)} + 3}}} \right)
Substituting f(9)=9f\left( 9 \right) = 9 and f′(9)=1f'\left( 9 \right) = 1 in the expression, we get
⇒lim⁡x→93−f(x)3−x=1×(9+39+3)\Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = 1 \times \left( {\dfrac{{\sqrt 9 + 3}}{{\sqrt 9 + 3}}} \right)
Simplifying the expression, we get
⇒lim⁡x→93−f(x)3−x=1×(1) ⇒lim⁡x→93−f(x)3−x=1\begin{array}{l} \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = 1 \times \left( 1 \right)\\\ \Rightarrow \mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} = 1\end{array}
Therefore, the value of the expression lim⁡x→93−f(x)3−x\mathop {\lim }\limits_{x \to 9} \dfrac{{3 - \sqrt {f\left( x \right)} }}{{3 - \sqrt x }} is 1.
Thus, the correct option is option (b).

Note: We rewrote the expression lim⁡x→9f(x)−f(9)x−9×(x+3f(x)+3)\mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} \times \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right) as lim⁡x→9f(x)−f(9)x−9×lim⁡x→9(x+3f(x)+3)\mathop {\lim }\limits_{x \to 9} \dfrac{{f\left( x \right) - f\left( 9 \right)}}{{x - 9}} \times \mathop {\lim }\limits_{x \to 9} \left( {\dfrac{{\sqrt x + 3}}{{\sqrt {f\left( x \right)} + 3}}} \right). This is because an expression of the form lim⁡x→af(x)g(x)\mathop {\lim }\limits_{x \to a} f\left( x \right)g\left( x \right) can be written as lim⁡x→af(x)×lim⁡x→ag(x)\mathop {\lim }\limits_{x \to a} f\left( x \right) \times \mathop {\lim }\limits_{x \to a} g\left( x \right). As the solution is lengthy, we might make a mistake in substituting the values correctly and therefore we need to be careful while solving. We will apply the limit at the end and not in the start because it will incur the wrong result. For example, in this question if we put the limit in the start then it will be equal to 0, which is wrong.