Question

How do you find the limit of following question $\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 6x + 5}}{{{x^2} - 25}}$?

Answer 1

We need to check the above expression whether it is in $\dfrac{0}{0}$ form or not.
If it is in $\dfrac{0}{0}$ form, then we will factorise the above expression to find the shortest form of it. On doing some simplification we get the required answer.

Formula Used:
Let say, $\mathop {\lim }\limits_{x \to a} \dfrac{m}{n}$ is an expression of limit$x$, which is tending to the value of $a$.
To simplify this type of expression, we need to see that the expression is in $\dfrac{0}{0}$ form or not.
We can further write the following iteration for the above expression:
$\mathop {\lim }\limits_{x \to a} \dfrac{m}{n} = \dfrac{{\mathop {\lim }\limits_{x \to a} m}}{{\mathop {\lim }\limits_{x \to a} n}}$.
Also, we need the following algebraic formula:
$({a^2} - {b^2}) = (a + b)(a - b).$

Complete step-by-step answer:
The given expression is:
$\mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 6x + 5}}{{{x^2} - 25}}$.
Let's say, the given expression is a function of $f(x)$.
So, we can write the following expression also:
$f(x) = \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 6x + 5}}{{{x^2} - 25}}$.
So, we have to put the value of $x$is equal to $5$, to check that the function of the numerator and the denominator is in $\dfrac{0}{0}$ form or not.
After putting$x = 5$, we get:
$\Rightarrow \dfrac{{{x^2} - 6x + 5}}{{{x^2} - 25}}$
$\Rightarrow \dfrac{{{5^2} - 6 \times 5 + 5}}{{{5^2} - 25}}$
Now, by simplifying the above iteration, we get:
$f(x) = \dfrac{{{5^2} - 6 \times 5 + 5}}{{{5^2} - 25}} = \dfrac{{25 - 30 + 5}}{{25 - 25}} = \dfrac{0}{0}.$
So, the given expression is in indeterminate form.
So, we need to further factorise the numerator and denominator.
After factorise the numerator, we get: ${x^2} - 6x + 5$
$\Rightarrow {x^2} - 5x - x + 5$
Taking the same term as common and we get
$\Rightarrow x(x - 5) - 1(x - 5)$
On rewriting we get
$\Rightarrow (x - 5)(x - 1).$
Now, by using the algebraic formula, we can further simplify the denominator as following:
$\Rightarrow {x^2} - 25 = {(x)^2} - {(5)^2} = (x + 5)(x - 5).$
Now, put the values of numerator and denominator, we get the following expression:
$f(x) = \mathop {\lim }\limits_{x \to 5} \dfrac{{{x^2} - 6x + 5}}{{{x^2} - 25}}$
$\Rightarrow \mathop {\lim }\limits_{x \to 5} \dfrac{{(x - 5)(x - 1)}}{{(x + 5)(x - 5)}}$.
Now, by cancel the common terms in numerator and the denominator, we get:
$f(x) = \mathop {\lim }\limits_{x \to 5} \dfrac{{(x - 1)}}{{(x + 5)}}.$
We can simplify it further as following:
$f(x) = \dfrac{{\mathop {\lim }\limits_{x \to 5} (x - 1)}}{{\mathop {\lim }\limits_{x \to 5} (x + 5)}}.$
As $x$ tends to $5$, we have to put this value to the above expression:
$f(x) = \dfrac{{(5 - 1)}}{{(5 + 5)}}.$
By simplifying it, we get:
$f(x) = \dfrac{4}{{10}} = \dfrac{2}{5}.$

$\therefore$The required value of the given expression is $\dfrac{2}{5}.$

Note:
Points to be remembered as follows:

$\dfrac{0}{0}$ is not only the indeterminate state,
$\dfrac{\infty }{\infty }$, $0.\infty$ and ${0^0}$ is also undefined form.