What is the limit of #((64x^2 +x)^(1/2) -8x)# as x approaches infinity?

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Answer 1 · 2016-05-23T14:33:54

#lim_(xrarroo)(sqrt(64x^2+x)-8x) = 1/16#

#lim_(xrarroo)(sqrt(64x^2+x)-8x)# has indeterminate initial form #oo-oo#.

We'll write this as a quotient using the conjugate of the expression.

#(sqrt(64x^2+x)-8x) = (sqrt(64x^2+x)-8x)/1*((sqrt(64x^2+x)+8x))/((sqrt(64x^2+x)+8x))#

# = (64x^2+x-64x^2)/(sqrt(64x^2+x)+8x)#

# = x/(sqrt(64x^2+x)+8x)#

For #x != 0#,, we can factor #x^2# in the radicand.

# = x/(sqrt(x^2(64+1/x))+8x)#

Now use #sqrt(x^2)=absx#, so for #x > 0#, we get

# = x/(sqrt(x^2)sqrt(64+1/x)+8x)= x/(xsqrt(64+1/x)+8x)#

Simplify:

# = x/(x(sqrt(64+1/x)+8)) = 1/(sqrt(64+1/x)+8)# (for #x > 0#)

Finally, evaluate the limit.

#lim_(xrarroo)1/(sqrt(64+1/x)+8) = 1/(sqrt(64+0)+8) = 1/16#