How do you find #\lim _ { x \rightarrow \infty } \frac { x ^ { 2} - 1} { 2x }#?

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Answer 1 · 2017-09-02T17:05:07

We can rewrite as

#L = lim_(x->oo) x^2/(2x) - 1/(2x)#

#L = lim_(x->oo) x - 1/(2x)#

We know that #lim_(x->oo) 1/x = 0# and #lim_(x->oo) x = oo#, so

#L = oo#

Hopefully this helps!

Answer 2 · 2017-09-02T17:43:58

#+oo#

Divide terms on numerator/denominator by the highest power of x that is #x^2#

#lim_(xtooo)(x^2/x^2-1/x^2)/((2x)/x^2)#

#=lim_(xtooo)(1-1/x^2)/(2/x)#

#=(1-0)/0#

#=+oo#