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Method of Difference

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Question

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

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Solution

$\begin{matrix} W e h a v e {lim}_{x \to \infty} \frac{x^{3} - 2 x^{2} + 3 x - 4}{4 x^{3} - 3 x^{2} + 2 x - 1} D i v i d e b o t h t h e n u m e r a t o r a n d d e n o m i t a t o r b y t h e h i g e s t d e g r e e o f x f o r t h i s q u e s t i o n t h e h i g e s t p o w e r i s x^{3} {lim}_{x \to \infty} \frac{\frac{x^{3}}{x^{3}} - \frac{2 x^{2}}{x^{3}} + \frac{3 x}{x^{3}} - \frac{4}{x^{3}}}{\frac{4 x^{3}}{x^{3}} - \frac{3 x^{2}}{x^{3}} + \frac{2 x}{x^{3}} - \frac{1}{x^{3}}} s i m p l i f y i n g w e h a v e {lim}_{x \to \infty} \frac{1 - \frac{2}{x} + \frac{3}{x^{2}} - \frac{4}{x^{3}}}{4 - \frac{3}{x} + \frac{2}{x^{2}} - \frac{1}{x^{3}}} w e k n o w t h a t {lim}_{x \to \infty} \frac{1}{x} = 0 T h e n a l l t h e t e r m s o t h e r t h a n t h e \frac{1}{4} c a n c l e t o 0. T h e r e f o r e, {lim}_{x \to \infty} \frac{x^{3} - 2 x^{2} + 3 x - 4}{4 x^{3} - 3 x^{2} + 2 x - 1} = \frac{1}{4} \end{matrix}$

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