How do you find the limit of #sqrt(9x^2 +x)-(3x)# as x approaches infinity?

Question

Christopher Agresta · Accepted Answer

# = 1/6#

#lim_(x to oo) sqrt(9x^2 +x)-3x#

#lim_(x to oo) 3x sqrt(1 +1/(9x))-3x#

#lim_(x to oo) 3x( (1 +1/(9x))^(1/2)-1)#

by Binomial Expansion

#lim_(x to oo) 3x( 1 + 1/2 1/(9x) + O(1/x^2) -1)#

#lim_(x to oo) 3x( 1/(18x) + O(1/x^2) )#

#lim_(x to oo) 1/6 + O(1/x) = 1/6#

Ezekiel Alexander · Answer

#1/6# #((sqrt(9x^2+x)-3x))/1 * ((sqrt(9x^2+x)-3x))/((sqrt(9x^2+x)-3x)) = (9x^2+x-9x^2)/((sqrt(9x^2+x)-3x))# # = x/(sqrt(9x^2+x)-3x)# # = x/(sqrt(x^2)sqrt(9+1/x) - 3x)# #" "# for #x != 0# # = x/(x(sqrt(9+1/x) - 3))# # = 1/(sqrt(9+1/x) - 3)# Now, as #x rarr oo#, #1/x rarr 0#, so the limit is #1/(sqrt9 + 3) = 1/6#

Isabella Ackerman · Answer

undefined To find the limit of sqrt(9x^2 + x) - (3x) as x approaches infinity, we can simplify the expression by factoring out x from the square root term. This gives us sqrt(x^2 * (9 + 1/x)) - (3x). Simplifying further, we have x * sqrt(9 + 1/x) - 3x.

Next, we can divide the entire expression by x to get sqrt(9 + 1/x) - 3.

As x approaches infinity, 1/x approaches 0. Therefore, the expression becomes sqrt(9 + 0) - 3, which simplifies to sqrt(9) - 3.

The square root of 9 is 3, so the final limit is 3 - 3, which equals 0.