lim (x approaches infinity) x/((x^2+4)^(1/2))
show steps please
show steps please
-
lim(x→∞) x / (x^2 + 4)^(1/2)
= lim(x→∞) 1 / [(1/2)(x^2 + 4)^(-1/2) * 2x], by L'Hopital's Rule
= lim(x→∞) (x^2 + 4)^(1/2) / x
Applying it again yields the original form of the limit!
-----------------
Try this instead:
Since lim(x→∞) x^2 / (x^2 + 4)
= lim(x→∞) 2x/(2x + 0), by L'Hopital's Rule
= 1,
lim(x→∞) x / (x^2 + 4)^(1/2)
= lim(x→∞) [x^2 / (x^2 + 4)]^(1/2)
= 1^(1/2)
= 1.
I hope this helps!
= lim(x→∞) 1 / [(1/2)(x^2 + 4)^(-1/2) * 2x], by L'Hopital's Rule
= lim(x→∞) (x^2 + 4)^(1/2) / x
Applying it again yields the original form of the limit!
-----------------
Try this instead:
Since lim(x→∞) x^2 / (x^2 + 4)
= lim(x→∞) 2x/(2x + 0), by L'Hopital's Rule
= 1,
lim(x→∞) x / (x^2 + 4)^(1/2)
= lim(x→∞) [x^2 / (x^2 + 4)]^(1/2)
= 1^(1/2)
= 1.
I hope this helps!