Question #567d4

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Answer 1 · 2018-01-24T17:15:07

#-oo#

#lim_(x to 0)[(1+2x)^1/2-1]/abs(x)#

#= ((2x-1)/2)/absx#

# = ((2x-1))/(2absx)#

#lim_(xrarra)[f(x)*g(x)]=lim_(xrarra)f(x) * lim_(xrarra)g(x)#

#= 1/2 *lim_(xrarr0)(2x-1)*lim_(xrarr0)1/absx#

#= 1/2 *lim_(xrarr0)(2*0-1)*lim_(xrarr0)1/absx#

graph{1/|x| [-9.25, 10.75, -0.92, 9.08]}

graph{1/|x| [-10.08, 9.92, -3.64, 6.36]}

#y = 1/|x|,x=x#

#lim_(x to 0)[(1+2x)^1/2-1]/abs(x) =1/2 * -1 * oo#

#lim_(x to 0)[(1+2x)^1/2-1]/abs(x) = -1 *oo#

#lim_(x to 0)[(1+2x)^1/2-1]/abs(x) = -oo#