Question

How do you find the limit of $\dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}}$ as $x \to -\infty$?

Answer 1

In order to find the solution of the given question that is to find the limit of $\dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}}$as $x \to -\infty$ divide the polynomial of highest degree ${{x}^{4}}$ from both numerator and denominator of the given expression. Then use the result to find the limit of given expression that is $\displaystyle \lim_{x \to -\infty }\left( \dfrac{1}{{{x}^{n}}} \right)=0$ for $n\ge 1$.

Complete step-by-step solution:
According to the question, given expression in the question is as follows:
$\dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}}$
To find the limit of the given expression, divide the polynomial of highest degree ${{x}^{4}}$ from both numerator and denominator of the given expression, we get:
$\Rightarrow \displaystyle \lim_{x \to -\infty }\left( \dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}} \right)=\displaystyle \lim_{x \to -\infty }\dfrac{\dfrac{\left( 1-2{{x}^{2}}-{{x}^{4}} \right)}{{{x}^{4}}}}{\dfrac{\left( 5+x-3{{x}^{4}} \right)}{{{x}^{4}}}}$
Now simplify the terms in the numerator and denominator by opening the bracket and dividing ${{x}^{4}}$ to each term, we get:
$\Rightarrow \displaystyle \lim_{x \to -\infty }\dfrac{\dfrac{1}{{{x}^{4}}}-\dfrac{2{{x}^{2}}}{{{x}^{4}}}-\dfrac{{{x}^{4}}}{{{x}^{4}}}}{\dfrac{5}{{{x}^{4}}}+\dfrac{x}{{{x}^{4}}}-\dfrac{3{{x}^{4}}}{{{x}^{4}}}}$
After this solve the terms with the help of the formula that is $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$, we will have:
$\Rightarrow \displaystyle \lim_{x \to -\infty }\dfrac{\dfrac{1}{{{x}^{4}}}-2{{x}^{2-4}}-1}{\dfrac{5}{{{x}^{4}}}+{{x}^{1-4}}-3}$
After simplifying the above expression, we get:
$\Rightarrow \displaystyle \lim_{x \to -\infty }\dfrac{\dfrac{1}{{{x}^{4}}}-2{{x}^{-2}}-1}{\dfrac{5}{{{x}^{4}}}+{{x}^{-3}}-3}$
Apply the formula ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}$ in the above expression, we will have:
$\Rightarrow \displaystyle \lim_{x \to -\infty }\dfrac{\dfrac{1}{{{x}^{4}}}-\dfrac{2}{{{x}^{2}}}-1}{\dfrac{5}{{{x}^{4}}}+\dfrac{1}{{{x}^{3}}}-3}$
Now apply the result $\displaystyle \lim_{x \to -\infty }\left( \dfrac{1}{{{x}^{n}}} \right)=0$ for $n\ge 1$ in the above expression, we will have:
$\Rightarrow \dfrac{0-2\left( 0 \right)-1}{5\left( 0 \right)+0-3}$
After simplifying the terms in numerator and denominator by opening the bracket, we get:
$\Rightarrow \dfrac{-1}{-3}$
As we can see minus is there in both numerator and denominator, so it will get cancelled and we will have:
$\Rightarrow \dfrac{1}{3}$
$\Rightarrow \displaystyle \lim_{x \to -\infty }\left( \dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}} \right)=\dfrac{1}{3}$
Therefore, limit of the expression $\dfrac{1-2{{x}^{2}}-{{x}^{4}}}{5+x-3{{x}^{4}}}$ is equal to $\dfrac{1}{3}$.

Note: In question like where we have to find the limit of rational function, there’s a standard result which is being used to solve this type of question that is if $f\left( x \right)=\dfrac{p\left( x \right)}{q\left( x \right)}=\dfrac{{{a}_{n}}{{x}^{n}}+{{a}_{n-1}}{{x}^{n-1}}+\cdot \cdot \cdot +{{a}_{2}}{{x}^{2}}+{{a}_{1}}{{x}^{1}}}{{{b}_{n}}{{x}^{n}}+{{b}_{n-1}}{{x}^{n-1}}+\cdot \cdot \cdot +{{b}_{2}}{{x}^{2}}+{{b}_{1}}{{x}^{1}}}$ then

& +\infty \text{ or }-\infty \text{ if }n >m \\\ & \dfrac{{{a}_{n}}}{{{b}_{n}}}\text{ if }n=m \\\ & 0\text{ if }n< m \\\ \end{aligned} \right.$$. In the above question, we have the 2nd case that is the power of the highest degree of the polynomial of numerator is equal to the power of the highest degree of the polynomial of the denominator.