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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#?

Answer 1

Samantha Adkison

# 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 #

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Answer 2

William Abdalla

To find the limit of g(x) as x approaches 1, we need to evaluate the function g(x) at x = 1. By substituting x = 1 into the given inequality, we have 2(1) ≤ g(1) ≤ 1^4 - 1^2 + 2. Simplifying this, we get 2 ≤ g(1) ≤ 2. Therefore, the limit of g(x) as x approaches 1 is 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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#?