Question

How do you find a value of k such that the limit $x \to 1$ exists given $\dfrac{{{x^2} - kx + 9}}{{x - 1}}$?

Answer 1

Hint: We will notice that the denominator becomes zero when we put in x = 1, therefore, we will somehow make the numerator zero as well so that we get 0/0 form and the limit exists.

Complete step by step solution:
We are given that the limit of the function given $\dfrac{{{x^2} - kx + 9}}{{x - 1}}$ exists where $x \to 1$. We need to find the value of k for such conditions.
Let us term the given function $\dfrac{{{x^2} - kx + 9}}{{x - 1}}$ as $f(x)$.
So, we have: $f(x) = \dfrac{{{x^2} - kx + 9}}{{x - 1}}$.
Now, we are given that $\mathop {\lim }\limits_{x \to 0} f(x)$ exists.
Therefore, $\mathop {\lim }\limits_{x \to 0} \left( {\dfrac{{{x^2} - kx + 9}}{{x - 1}}} \right)$ exists.
Now, since if we put in $x = 1$, we get the denominator of $f(1)$ to be equal to 0.
Therefore, we need to get $\dfrac{0}{0}$ form, so that we can use L – Hospital’s rule.
Therefore, the numerator of $f(x)$ must be equal to 0.
Therefore, we have ${x^2} - kx + 9 = 0$.
Using the quadratic formula, we have:-
$ \Rightarrow x = \dfrac{{ - ( - k) \pm \sqrt {{{( - k)}^2} - 4 \times 1 \times 9} }}{{2 \times 1}}$
Simplifying the calculations in the above equation, we will then obtain the following equation:-
$ \Rightarrow x = \dfrac{{k \pm \sqrt {{k^2} - 36} }}{2}$
Putting $x = 1$ in the above equation, we will then get the following equation:-
$ \Rightarrow 2 = \dfrac{{k \pm \sqrt {{k^2} - 36} }}{2}$
Multiplying both the sides of equation by 2, we will get:-
$ \Rightarrow k \pm \sqrt {{k^2} - 36} = 2$
Taking k from addition in LHS to subtraction in RHS, we will then obtain the following equation:-
$ \Rightarrow \pm \sqrt {{k^2} - 36} = 2 - k$
Taking square of both sides of above equation, we have:-
$ \Rightarrow {k^2} - 36 = {\left( {2 - k} \right)^2}$
Now, we know that we have an identity given by ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.
Therefore, we have:-
$ \Rightarrow {k^2} - 36 = 4 + {k^2} - 4k$
Crossing off ${k^2}$ from both the sides of the above equation, we have:-
$ \Rightarrow - 36 = 4 - 4k$
Subtracting 4 from both the sides, we have:-
$ \Rightarrow - 4k = - 40$

Hence, k = 10 is the required answer

Note:
The students must note that this could have been done in a simpler way as well without the hassle of quadratic formula and everything as well as follows:-
Since the denominator of the given function is zero when x is 1, therefore, we will put the numerator of f (1) to be zero as well.
So, ${1^2} - k \times 1 + 9 = 0$
Thus, we have k = 10.