How do you find #lim sqrt(x^2+1)-x# as #x-oo#?

Question

How do you find #lim sqrt(x^2+1)-x# as #x->oo#?

Paisley Johnson · Accepted Answer

Let #"f"(x) = (x^2+1)^(1/2) - x#.

Then we can write,

#"f"(x) = (x^2* (1 + 1/x^2))^(1/2)-x# #"f"(x) = abs(x)*(1+1/x^2)^(1/2)-x#.

Then, we can expand #(1+1/x^2)^(1/2)# using the binomial theorem for #abs(1/x^2)<1#.

#abs(1/x^2)<1# for #1/x^2<1# as #1/x^2# is always positive. #1/x^2<1# for #x^2>1#, so #x>1# or #x<-1#, i.e. #abs(x) > 1#.

Then for #abs(x)>1#,

#"f"(x) = abs(x) * (1+(1/2)*1/x^2+(1/2)(-1/2)*1/x^4+...) - x# #"f"(x) = abs(x)-x + 1/2*abs(x)/x^2+(1/2)(-1/2)*abs(x)/x^4+...#

Clearly #lim_{x\to\infty}abs(x)/x^n = 0# for #n>1#.

Then, as this is expansion is valid for any #x>1#,

#lim_{x\to\infty} "f"(x) = abs(x)-x#.

#abs(x)={ (x, x>=0), (-x, x<=0) :}#.

Then, we conclude,

#lim_{x\to+\infty} "f"(x) = 0#, #lim_{x\to-\infty} "f"(x) = +\infty#.

Ezra Adams · Answer

#lim_(xrarroo)(sqrt(x^2+1)-x)# takes indeterminate form #oo-oo# #(sqrt(x^2+1)-x)/1 = ((sqrt(x^2+1)-x))/1 * ((sqrt(x^2+1)+x))/((sqrt(x^2+1)+x))# # = (x^2+1-x^2)/(sqrt(x^2+1)+x)# # = 1/(sqrt(x^2+1)+x)# Now use 
#sqrt(x^2+1) = sqrt(x^2)sqrt(1+1/x^2)# for all #x != 0# # = abs(x)sqrt(1+1/x^2)# for all #x != 0# # = xsqrt(1+1/x^2)# for all #x > 0# (We want #lim_(xrarroo)#) Returning to  #(sqrt(x^2+1)-x) =  1/(sqrt(x^2+1)+x)# # = 1/(xsqrt(1+1/x^2))# As #xrarroo#, we get a limit of #0# Bonus answer Ar #xrarr-oo#, the expression #sqrt(x^2+1)-x# takes non-indeterminate form #oo-(-oo) = oo+oo=oo#

Emma Aldridge · Answer

undefined To find the limit of sqrt(x^2+1)-x as x approaches infinity, we can simplify the expression. By multiplying the numerator and denominator by the conjugate of the expression, we can eliminate the square root:

lim sqrt(x^2+1)-x as x->oo
= lim ((sqrt(x^2+1)-x) * (sqrt(x^2+1)+x)) / (sqrt(x^2+1)+x) as x->oo
= lim (x^2+1 - x^2) / (sqrt(x^2+1)+x) as x->oo
= lim 1 / (sqrt(x^2+1)+x) as x->oo

As x approaches infinity, the denominator also approaches infinity. Therefore, the limit is:

lim 1 / (sqrt(x^2+1)+x) as x->oo = 0

How do you find #lim sqrt(x^2+1)-x# as #x-&gt;oo#?

How do you find #lim sqrt(x^2+1)-x# as #x->oo#?