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Question Number 197301 by uchihayahia last updated on 13/Sep/23

h o w d o i c a l c u l a t e t h i s

\lim_{x \to - \infty} \frac{x^{4} + 2 x^{2} + x - 2}{x^{3} + 2 x^{2} + x - 1}

m u l t i p l y i n g b o t h n u m e r a t o r

a n d d e n u m e r a t o r b y \frac{1}{x^{4}}

\lim_{x \to - \infty} \frac{1 + \frac{2}{x^{2}} + \frac{1}{x^{3}} - \frac{2}{x^{4}}}{\frac{1}{x} + \frac{2}{x^{2}} + \frac{1}{x^{3}} - \frac{1}{x^{4}}}

\frac{1 + 0 + 0 - 0}{0 + 0 + 0 - 0}

\infty

w h i c h i s n o t t r u e t h e a n s w e r i s - \infty,

i t r i e d m u l t i p l y i n g \frac{1}{x^{3}} a n d g o t - \infty

b u t s t i l l c o n f u s e d w h a t d i d i d o w r o n g

u s i n g \frac{1}{x^{4}}

Answered by AST last updated on 13/Sep/23

\frac{x (x^{3} + 2 x^{2} + x - 1) - 2 x^{3} + x^{2} + 2 x - 2}{x^{3} + 2 x^{2} + x - 1}

= x - \frac{2 x^{3} + 4 x^{2} + 2 x - 2 - 5 x^{2} - 4 x + 4}{x^{3} + 2 x^{2} + x - 1}

= x - 2 - \frac{5 x^{2} + 4 x - 4}{x^{3} + 2 x^{2} + x - 1} = x - 2 - \frac{\frac{5}{x} + \frac{4}{x^{2}} - \frac{4}{x^{3}}}{1 + \frac{2}{x} + \frac{1}{x^{2}} - \frac{1}{x^{3}}}

\to - \infty

Commented by uchihayahia last updated on 16/Sep/23

t h a n k y o u, i a l w a y s t h o u g h t i s h o u l d

d i v i d e b o t h b y x^{n}, m i s h i g h e s t d e g r e e