Question

the value of Lt
$x - > 0 \sqrt{1 + x + { x}^{2} } \div x$
find the limit​

Answer 1

Answer:

$\lim _{ x \to0 } \frac{ \sqrt{1 + x + {x}^{2} } }{x} \\ \\ = { \frac{ \sqrt{1 + 0 + {0}^{2} }}{0} } \\ \\ = \frac{1}{0} \\ = \infty$

Which is not undetermined form . So limit doesn't exist