$\lim _{x\rightarrow \dfrac {1}{2}}\dfrac {8x^{3}-1}{16x^{4}-1}$ limitinin değeri kaçtır?
Cevap $3/4$
benim denediğim çözüm
$\lim _{x\rightarrow \dfrac {1}{2}}\dfrac {8\cdot \dfrac {1}{8}-1}{16\cdot \dfrac {1}{16}-1}=\dfrac {0}{0}$ belirsizliği
$\dfrac {\left( 2x\right) ^{3}-1^{3}}{\left( 2x\right) ^{4}-1^{4}}$
$=\dfrac {\left( 2x-1\right) \left( 4x^{2}+2x+1\right) }{\left( 4x^{2}+1\right) \left( 2x+1\right) \left( 2x-1\right) }$
$=\dfrac {4x^{2}+2x+1}{\left( 4x^{2}+1\right) \left( 2x+1\right) }$
$=\dfrac {4\cdot \dfrac {1}{4}+2\cdot \dfrac {1}{2}+1}{\left( \cdot 4\cdot \dfrac {1}{4}+1\right) \left( 2\cdot \dfrac {1}{2}+1\right) }$
$=\dfrac {1+1+1}{\left( 1+1\right) \left( 1+1\right) }$$=3/4$