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

Question

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

Isaiah Allen · Accepted Answer

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

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.

Christopher Allers · Answer

undefined To find the limit of (x+1)^(3/2) - x^(3/2) as x approaches infinity, we can simplify the expression by using the binomial expansion. The binomial expansion of (x+1)^(3/2) is given by:

(x+1)^(3/2) = x^(3/2) + (3/2)x^(1/2) + (3/8)x^(-1/2) + ...

As x approaches infinity, the higher-order terms become negligible compared to the dominant term x^(3/2). Therefore, we can ignore the terms (3/2)x^(1/2), (3/8)x^(-1/2), and so on.

Thus, the expression simplifies to:

(x+1)^(3/2) - x^(3/2) ≈ x^(3/2) - x^(3/2) = 0

Therefore, the limit of (x+1)^(3/2) - x^(3/2) as x approaches infinity is 0.

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

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