How do you find #lim sqrt(x^2+1)-x# as #x->oo#?
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"How does implicit differentiation work?"
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#.
#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#
