Question

lim x->infinity lnx/sqrt infinity

Answer 1

wirte the function clearly

Answer 2

use L' Hospital rule ...ok

Answer 3

sqrt of what?

Answer 4

\[\lim_{x \rightarrow \infty} \frac{ lnx }{ \sqrt{x} }\]

Answer 5

i m sorry but i think that u r given orders
by this prase "use L' Hospital rule ...ok"
sorry maybe another person
sorry

Answer 6

its okay

Answer 7

Using L'Hopital rule,
\[\lim_{x\rightarrow∞}\frac{1/x}{1/(2\sqrt{x})}=\lim_{x\rightarrow∞}\frac{2}{\sqrt{x}}=0\]

Answer 8

or use your eyeballs
log grows slower than any power of \(x\)

Answer 9

when u plug in the \[\infty \ -> \sqrt{x}\] y is it 0

Answer 10

Yes, the radical function grows faster than a constant :)

Answer 11

wat?

Answer 12

im asking when u plug in infinity in sqrtx why is the answer 0

Answer 13

the log grows very very slowly so
\[\lim_{x\to \infty}\frac{\ln(x)}{x^p}=0\] so long as \(p>0\)

Answer 14

Well, it is 0, because the function Square root x, grows, while the number 2 is constant, so if you do an easy calculus in a calculator, you will see that the limit is 0. Also you can check proofs with epsilon delta formalism.

Answer 15

don't get it

Answer 16

If x is large, then you can put in the calculator x=10000,
\[2/\sqrt{10000}=0.02\]
\[2/\sqrt{1000000}=0.002\]
So if x increases, the function "tends" to 0.

Answer 17

do u mean when u plug in infinity it gets closer to zero that's y its zero

Answer 18

Yes, right.

Answer 19

oh ok thank u :)

Answer 20

y is zero because the numerator is growing slower than the denominator. not because you plug in infinity.