How do you evaluate $\dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4}$ as $x$ approaches $-4$ ?

Hint: In this question we have been asked to evaluate the given expression $\dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4}$ when $x$ approaches $-4$ . For that we will substitute $-4$ in place of $x$ and then evaluate the value. This is a case from limits.

Complete step by step solution:
Now considering from the question we need to evaluate the given expression $\dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4}$ at $x\to -4$ which implies that we have $\displaystyle \lim_{x\to -4}\left( \dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4} \right)$ .
When we substitute $x=-4$ in the expression $\dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4}$ then we will have $\Rightarrow \dfrac{{{\left( -4 \right)}^{2}}+5\left( -4 \right)+4}{{{\left( -4 \right)}^{2}}+3\left( -4 \right)-4}=\dfrac{16-20+4}{16-12-4}\Rightarrow \dfrac{0}{0}$ .
Now as it is in the undefined form. We will use the L’Hospital rule which states that when the expression is in an undefined form then the limit is evaluated by differentiating the numerator and denominator separately with respect to the variable $x$ and replacing in the expression.
Now we will differentiate the numerator after differentiating we will have $\dfrac{d}{dx}\left( {{x}^{2}}+5x+4 \right)=2x+5$ .
Now we will differentiate the denominator after differentiating we will have $\dfrac{d}{dx}\left( {{x}^{2}}+3x-4 \right)=2x+3$ .
Now we will replace the expressions in the function.
After that we will have $\Rightarrow \dfrac{2x+5}{2x+3}$ .
Now we will again replace $x$ with $-4$ then we will have $\Rightarrow \dfrac{2\left( -4 \right)+5}{2\left( -4 \right)+3}=\dfrac{-8+5}{-8+3}\Rightarrow \dfrac{-3}{-5}=\dfrac{3}{5}$

Therefore we can conclude that the evaluation of the expression $\dfrac{\left( {{x}^{2}} \right)+5x+4}{\left( {{x}^{2}} \right)+3x-4}$ as $x$ approaches $-4$ gives the value $\dfrac{3}{5}$ .

Note: During solving this question we have to be very careful if we do not apply the concept of L’Hospital then we will be unable to solve the answer. Similarly we can evaluate any expressions for example if we consider the expression $\dfrac{{{x}^{2}}+2x}{{{x}^{2}}+5x}$ then we will have the evaluation result will be $\dfrac{0}{0}$ when $x$ approaches to zero. By applying L’Hospital rule we will have $\Rightarrow \dfrac{2x+2}{2x+5}=\dfrac{2}{5}$ .