Question

=> Don't Spam
=> Right Answer=Brainliest

EVALUATE

$\huge{ \boxed { \boxed{\bold \red{ \displaystyle \lim_{x \to 1} - \dfrac{x - 4}{ {x}^{2} - 6x + 8 }}}}}$​

Answer 1

Solution:

First let's understand the concept applied in solving this question.

❒ Concept used:

✯ In such questions substitute the value where the limit is tending to in the expression. Then we will get the value of this expression.

➵ But note that if the value after substitution comes to be in the form of 0/0 or ∞/∞ form, then apply L'hospital rule.

★ Step-by-step explanation:

$\boldsymbol{ \lim_{{x} \to 1 \:} - \bigg( \dfrac{x - 4}{ {x}^{2} - 6x + 8 }} \bigg)$

On substituting the limits,

$\implies \sf{ - \bigg( \dfrac{1 - 4}{ {1}^{2} - 6(1) + 8 } \bigg) }$

$\implies \sf{ - \bigg( \dfrac{ - 3}{1 - 6 + 8} \bigg)}$

$\implies \sf{ - \bigg( \dfrac{ - 3}{3} \bigg) }$

$\implies \sf{ - ( - 1)}$

$= 1$

$\therefore \: \boxed{ \bf{ \lim_{x \to 1} - \bigg( \dfrac{x - 4}{ {x}^{2} - 6x + 8 } \bigg) = 1}}$

Answer 2

$\bold \red{ \displaystyle \lim_{x \to 1} - \dfrac{x - 4}{ {x}^{2} - 6x + 8 }}$

$\bold \red{ \displaystyle \lim_{x \to 1} \dfrac{4 - x}{ {x}^{2} - 6x + 8 }}$

Substitute the variable with the value:

$\bold \red{ \displaystyle \lim_{x \to 1} \dfrac{4 - x}{ {x}^{2} - 6x + 8 }} = 1$

Therefore,

$\bold \red{ \displaystyle \lim_{x \to 1} \dfrac{4 - x}{ {x}^{2} - 6x + 8 }} = 1$

Answer:

$\bold \red{ \displaystyle \lim_{x \to 1} \dfrac{4 - x}{ {x}^{2} - 6x + 8 }} = 1$