How do you find the limit #(1+5/sqrtx)/(2+1/sqrtx)# as #x->0^+#?

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Answer 1 · 2016-11-04T13:53:07

# lim_(x->0^+)((1+5/sqrt(x))/(2+1/sqrt(x))) = 5#

If we look at the graph of #y=(1+5/sqrt(x))/(2+1/sqrt(x))# we can see that it is clear that the limit exists, and is approximately #5#

graph{(1+5/sqrt(x))/(2+1/sqrt(x)) [-8, 8, -2, 10]}

Now, As #x->0# then #1/x->oo# and #1/sqrtx->oo# but if we can invert these expressions they both #->0#

So, we look for a way to invert the #1/A# expression

# lim_(x->0^+)((1+5/sqrt(x))/(2+1/sqrt(x))) = lim_(x->0^+)sqrtx/sqrtx * (1+5/sqrt(x))/(2+1/sqrt(x)) #

# :. lim_(x->0^+)((1+5/sqrt(x))/(2+1/sqrt(x))) = ((lim_(x->0^+)sqrtx)+5) / ((lim_(x->0^+)sqrtx)+1)#

# :. lim_(x->0^+)((1+5/sqrt(x))/(2+1/sqrt(x))) = (0+5)/(0+1)#

# :. lim_(x->0^+)((1+5/sqrt(x))/(2+1/sqrt(x))) = 5#

Which is completely consistent with the above graph.