How do you find #lim (x+1)^(3/2)-x^(3/2)# as #x->oo#?

1 Answer
Parabola
Jan 17, 2018

#oo# or the graph of the function will get undoubtedly large.

Explanation:

So we have #lim_ (x->oo)(x+1)^(3/2)-x^(3/2)#

We can see that this function is continuous when #x>=0#
Remember that when a function is continuous at #c#, then #lim_ (x->c)f(x)=f(c)#
So let's substitute #oo# in the place of #x#.

#(oo+1)^(3/2)-oo^(3/2)#

Now how do we solve that?

Well, let's use logic here.
If there is this really,really large number, and we are raising it to a power greater than one, will get an answer even greater than what we started with. Also, this function will give a positive value for any #x# values that are equal to or greater than one.

Therefore, our function will get undoubtedly large as #x# approaches infinity.

So you can say that #lim_ (x->oo)(x+1)^(3/2)-x^(3/2)# is #oo# or that it gets undoubtedly large.

We can even look at the graph of our function.
graph{(x+1)^(3/2)-x^(3/2) [-10, 10, -5, 5]}
Even though the rate that this is increasing is decreasing, there is no limit of how much the #y# value can be.

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