Question

Find the limit.\\ $\lim_{h \to 0^+} \frac{\sqrt{h^2 + 6h + 13} - \sqrt{13}}{h}$\\ $\lim_{h \to 0^+} \frac{\sqrt{h^2 + 6h + 13} - \sqrt{13}}{h} = \boxed{}$

          Find the limit.\\
$\lim_{h \to 0^+} \frac{\sqrt{h^2 + 6h + 13} - \sqrt{13}}{h}$\\
$\lim_{h \to 0^+} \frac{\sqrt{h^2 + 6h + 13} - \sqrt{13}}{h} = \boxed{}$

Find the limit.

limh → 0^+(√(h^2 + 6h + 13) - √(13))/(h)

limh → 0^+(√(h^2 + 6h + 13) - √(13))/(h) =

Added by Melissa M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Thanks for your feedback!

Find the limit. . lim_(h->0^(+))(sqrt(h^(2)+6h+13)-sqrt(13))/(h) lim_(h->0^(+))(sqrt(h^(2)+6h+13)-sqrt(13))/(h)= Find the limit. KK h+6h+13-13 lim h h-o* h+6h+13-13 lim h 04