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

Hint: 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.