For all #x>=0# and #2x<=g(x)<=x^4-x^2+2# how do you find the limit of g(x) as #x->1#?

Question

4240 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T01:45:53

# lim_(x rarr 1) g(x) =2 #

We can use the squeeze theorem (or the sandwich theorem), which basically states that, If:

# g(x) le f(x) le h(x) # and # lim_(x rarr a) g(x) = lim_(x rarr a) h(x) = L#

Then:

# lim_(x rarr a) f(x) = L #

So for this problem we have:

# 2x le g(x) le x^4-x^2+2 #

And so if we take the limit as #x rarr 1#; then

# lim_(x rarr 1) {2x} =2 #
# lim_(x rarr 1){x^4-x^2+2} =2 #

And so we can aply the squeeze theorem; which gives

# lim_(x rarr 1) g(x) =2 #