Question

precise limit definition (use definition and prove it). Check my work?

Answer 1

\[\lim_{x \rightarrow 2^-}\frac{ -1 }{ \sqrt{2x-4} }=-\infty\]

Answer 2

Definition: \[\forall M > 0 \exists \delta > 0 \in 0 < 2 - x < \delta, \frac{ -1 }{ \sqrt{2x-4} } < - M \]

Prove: \[\frac{ -1 }{ \sqrt{2x-4} } < - M\]
\[\frac{ 1 }{ M } > \sqrt{2x-4}\]
\[\frac{ 1 }{ M^2 } > 2x-4\]
\[\delta = \frac{ 1 }{ M^2 }\]
\[\forall M > 0 \exists \delta ( = \frac{ 1 }{ M^2 }) >0 \in 0 < 2 - x < \delta, \frac{ -1 }{ \sqrt{2x-4} } < - M\]
Thus, by definition, \[\lim_{x \rightarrow 2^-}\frac{ -1 }{ \sqrt{2x-4} }= -\infty \]