
How do you evaluate the limit \[\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{10-2{{x}^{4}}}\] as x approaches $\infty $ ?
Answer
557.7k+ views
Hint: Divide each term of the expression by the greatest power of the variable ‘x’ i.e. ${{x}^{4}}$. This can be done by multiplying the numerator and denominator by the reciprocal of the greatest power of ‘x’ i.e. $\dfrac{1}{{{x}^{4}}}$. Do the necessary simplification to bring the expression in the form of $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}$ etc. Put ‘0’ in place of $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}$ etc. when ‘x’ approaches $\infty $. The value of the expression can be obtained by further simplification.
Complete step by step answer:
As we know $\dfrac{1}{0}=\infty $
So, it can be written that $\dfrac{1}{\infty }=0$
Now if $x \to \infty $, then $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}\to 0$
Considering our expression
\[\displaystyle \lim_{x \to \infty }\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{10-2{{x}^{4}}}\]
The greatest power of ‘x’ in the expression is ${{x}^{4}}$
So, we need to divide it with each and every term of the expression.
The reciprocal of ${{x}^{4}}=\dfrac{1}{{{x}^{4}}}$
Hence, multiplying the numerator and denominator by $\dfrac{1}{{{x}^{4}}}$, we get
\[\begin{align}
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{10-2{{x}^{4}}}\cdot \dfrac{\dfrac{1}{{{x}^{4}}}}{\dfrac{1}{{{x}^{4}}}} \\
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{{{x}^{4}}}}{\dfrac{10-2{{x}^{4}}}{{{x}^{4}}}} \\
\end{align}\]
Dividing \[{{x}^{4}}\] with each and every term, we get
$\begin{align}
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{\dfrac{3{{x}^{4}}}{{{x}^{4}}}-\dfrac{{{x}^{2}}}{{{x}^{4}}}+\dfrac{5}{{{x}^{4}}}}{\dfrac{10}{{{x}^{4}}}-\dfrac{2{{x}^{4}}}{{{x}^{4}}}} \\
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{3-\dfrac{1}{{{x}^{2}}}+5\cdot \dfrac{1}{{{x}^{4}}}}{10\cdot \dfrac{1}{{{x}^{4}}}-2} \\
\end{align}$
As $x \to \infty $, so $\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{4}}}\to 0$
$\begin{align}
& \Rightarrow \dfrac{3-0+5\times 0}{10\times 0-2} \\
& \Rightarrow \dfrac{3+0}{0-2} \\
& \Rightarrow -\dfrac{3}{2} \\
\end{align}$
This is the required solution of the given expression.
Note:
The first approach to solve such a question should be, each term should be divided by the greatest power of the present variable. So, the numerator and denominator should be multiplied with the reciprocal of the greatest power of ‘x’ to bring the expression to a form where we can put $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}\to 0$ while $x \to \infty $. From the above solution it can be concluded that for a large ‘x’, the function is approaching a horizontal asymptote $y=-\dfrac{3}{2}$ .
Complete step by step answer:
As we know $\dfrac{1}{0}=\infty $
So, it can be written that $\dfrac{1}{\infty }=0$
Now if $x \to \infty $, then $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}\to 0$
Considering our expression
\[\displaystyle \lim_{x \to \infty }\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{10-2{{x}^{4}}}\]
The greatest power of ‘x’ in the expression is ${{x}^{4}}$
So, we need to divide it with each and every term of the expression.
The reciprocal of ${{x}^{4}}=\dfrac{1}{{{x}^{4}}}$
Hence, multiplying the numerator and denominator by $\dfrac{1}{{{x}^{4}}}$, we get
\[\begin{align}
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{10-2{{x}^{4}}}\cdot \dfrac{\dfrac{1}{{{x}^{4}}}}{\dfrac{1}{{{x}^{4}}}} \\
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{\dfrac{3{{x}^{4}}-{{x}^{2}}+5}{{{x}^{4}}}}{\dfrac{10-2{{x}^{4}}}{{{x}^{4}}}} \\
\end{align}\]
Dividing \[{{x}^{4}}\] with each and every term, we get
$\begin{align}
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{\dfrac{3{{x}^{4}}}{{{x}^{4}}}-\dfrac{{{x}^{2}}}{{{x}^{4}}}+\dfrac{5}{{{x}^{4}}}}{\dfrac{10}{{{x}^{4}}}-\dfrac{2{{x}^{4}}}{{{x}^{4}}}} \\
& \Rightarrow \displaystyle \lim_{x \to \infty }\dfrac{3-\dfrac{1}{{{x}^{2}}}+5\cdot \dfrac{1}{{{x}^{4}}}}{10\cdot \dfrac{1}{{{x}^{4}}}-2} \\
\end{align}$
As $x \to \infty $, so $\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{4}}}\to 0$
$\begin{align}
& \Rightarrow \dfrac{3-0+5\times 0}{10\times 0-2} \\
& \Rightarrow \dfrac{3+0}{0-2} \\
& \Rightarrow -\dfrac{3}{2} \\
\end{align}$
This is the required solution of the given expression.
Note:
The first approach to solve such a question should be, each term should be divided by the greatest power of the present variable. So, the numerator and denominator should be multiplied with the reciprocal of the greatest power of ‘x’ to bring the expression to a form where we can put $\dfrac{1}{x},\dfrac{1}{{{x}^{2}}},\dfrac{1}{{{x}^{3}}}\to 0$ while $x \to \infty $. From the above solution it can be concluded that for a large ‘x’, the function is approaching a horizontal asymptote $y=-\dfrac{3}{2}$ .
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