Odpowiedź:
[tex]\displaystyle\lim_{x \to -4^+} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{(-8)\cdot 0^+} =\frac{26}{0^-}= -\infty\\\mbox { asymptota pionowa x=-4}\\\lim_{x \to 4^-} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{0^-\cdot 8}=-\infty\\\lim_{x \to 4^+} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{0^+\cdot 8}=+\infty\\\mbox{asymptota pionowa x=4}[/tex]
[tex]\displaystyle \lim_{x \to \pm\infty} \frac{x^{2} +10/:x^{2} }{x^{2} -16/:x^{2} } =\frac{1+(\frac{10}{x^{2} } )^\nearrow^0}{1-(\frac{16}{x^{2} })^\nearrow^0 } =1\\\mbox{asymptota pozioma obustronna y=1}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Odpowiedź:
[tex]\displaystyle\lim_{x \to -4^+} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{(-8)\cdot 0^+} =\frac{26}{0^-}= -\infty\\\mbox { asymptota pionowa x=-4}\\\lim_{x \to 4^-} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{0^-\cdot 8}=-\infty\\\lim_{x \to 4^+} \frac{x^{2}+10 }{(x-4)(x+4)} =\frac{26}{0^+\cdot 8}=+\infty\\\mbox{asymptota pionowa x=4}[/tex]
[tex]\displaystyle \lim_{x \to \pm\infty} \frac{x^{2} +10/:x^{2} }{x^{2} -16/:x^{2} } =\frac{1+(\frac{10}{x^{2} } )^\nearrow^0}{1-(\frac{16}{x^{2} })^\nearrow^0 } =1\\\mbox{asymptota pozioma obustronna y=1}[/tex]