How do you find #lim 1/sqrtx-1/x# as #x->0^+#? Calculus Limits Limits at Infinity and Horizontal Asymptotes 1 Answer Andrea S. Aug 1, 2017 Simplify the expression: #1/sqrtx - 1/x = (x-sqrtx)/(xsqrtx) = (sqrtx-1)/x# Passing to the limit: #lim_(x->oo) (1/sqrtx - 1/x ) = lim_(x->oo) (sqrtx-1)/x = -1 xx +oo = -oo# Answer link Related questions What kind of functions have horizontal asymptotes? How do you find horizontal asymptotes for #f(x) = arctan(x)# ? How do you find the horizontal asymptote of a curve? How do you find the horizontal asymptote of the graph of #y=(-2x^6+5x+8)/(8x^6+6x+5)# ? How do you find the horizontal asymptote of the graph of #y=(-4x^6+6x+3)/(8x^6+9x+3)# ? How do you find the horizontal asymptote of the graph of y=3x^6-7x+10/8x^5+9x+10? How do you find the horizontal asymptote of the graph of #y=6x^2# ? How can i find horizontal asymptote? How do you find horizontal asymptotes using limits? What are all horizontal asymptotes of the graph #y=(5+2^x)/(1-2^x)# ? See all questions in Limits at Infinity and Horizontal Asymptotes Impact of this question 1370 views around the world You can reuse this answer Creative Commons License