How do you find #lim t(sqrt(t+1)-sqrtt)# as #t->oo#?

2 Answers
George C.
Jan 7, 2018

#lim_(t->oo) t(sqrt(t+1)-sqrt(t)) = oo#

Explanation:

#lim_(t->oo) t(sqrt(t+1)-sqrt(t))#

#=lim_(t->oo) (t(sqrt(t+1)-sqrt(t))(sqrt(t+1)+sqrt(t)))/(sqrt(t+1)+sqrt(t))#

#=lim_(t->oo) (t((t+1)-t))/(sqrt(t+1)+sqrt(t))#

#=lim_(t->oo) t/(sqrt(t+1)+sqrt(t))#

#=lim_(t->oo) sqrt(t) * sqrt(t)/(sqrt(t+1)+sqrt(t))#

#=lim_(t->oo) sqrt(t) * 1/(sqrt(1+1/t)+1)#

#=lim_(t->oo) 1/2sqrt(t)#

#=oo#

Jim S
Jan 7, 2018

#+oo#

Explanation:

I'll replace #t# with #x#

#lim_(xrarr+oo)x*(sqrt(x+1)-sqrtx)# #=#

#lim_(xrarr+oo)x*((sqrt(x+1)-sqrtx)(sqrt(x+1)+sqrtx))/(sqrt(x+1)+sqrtx)# #=#

#lim_(xrarr+oo)x*(sqrt(x+1)^2-sqrtx^2)/(sqrt(x+1)+sqrtx)# #=#

#lim_(xrarr+oo)x*(cancel(x)+1-cancel(x))/(sqrt(x+1)+sqrtx)# #=#

#lim_(xrarr+oo)x/(sqrt(x+1)+sqrtx)# #=#

#lim_(xrarr+oo)x/(sqrt(x^2(1/x+1/x^2))+sqrt(x^2*(1/x)))# #=#

#lim_(xrarr+oo)x/(|x|sqrt(1/x+1/x^2)+|x|sqrt(1/x))#

#x->+oo# , #x>0#

#=# #lim_(xrarr+oo)x/(xsqrt(1/x+1/x^2)+xsqrt(1/x))# #=#

#lim_(xrarr+oo)cancel(x)/(cancel(x)(sqrt(1/x+1/x^2)+sqrt(1/x))# #=#

#lim_(xrarr+oo)1/(sqrt(1/x+1/x^2)+sqrt(1/x))# #=# #+oo#

because #lim_(xrarr+oo)x=+oo# so #lim_(xrarr+oo)1/x=0#

and #lim_(xrarr+oo)sqrt(1/x)=0#

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