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Mar 26, 2018

Answer:

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = 1/4#

Explanation:

Substitute #x= cost#, #t=arccosx# so that:

#lim_(x->1^-) t(x) = arccos(1) = 0#

Then:

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = lim_(t->0^+) (1-sqrtcost)/t^2#

Rationalize the numerator:

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = lim_(t->0^+) ((1-sqrtcost)(1+sqrtcost))/(t^2(1+sqrtcost))#

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = lim_(t->0^+) (1-cost)/(t^2(1+sqrtcost))#

Use now the well known trigonometric limit:

#lim_(theta->0) (1-costheta)/theta^2 = 1/2#

to have:

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = lim_(t->0^+) (1-cost)/t^2 * lim_(t->0^+) 1/(1+sqrtcost)#

#lim_(x->1^-) (1-sqrtx)/(arccosx)^2 = 1/2 * 1/2 = 1/4#

graph{ (1-sqrtx)/(arccosx)^2 [-0.51, 1.99, -0.625, 0.625]}

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