How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x-0#?

Question

How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x->0#?

Ezra Adams · Accepted Answer

#lim_(x->0) (5x^2)/(2x) = 0#

The first line of attack when looking at a limit should be to plug in the number we want the function to approach.

In this case we get a weird #0/0# result. This seems to be in contradiction because the numerator says our limit is going to zero, but our denominator says it's going to infinity

This is an indeterminate form !

We solve indeterminate forms by using L'Hôpital's Rule.

If you're new to calculus, this might look a little messy, but I'll try to keep it simple, so bear with me. You can even skip the mumbo-jumbo and just read the math part if you don't need the theory.

L'Hôpital's Rule says that if #f# and #g# are differentiable functions and #g'(x)# does not equal zero on an interval containing #a#, and

#lim_(x->a) f(x) = 0# and #lim_(x->a) g(x) = 0#, or

#lim_(x->a) f(x) = +-oo# and #lim_(x->a) g(x) = +-oo#

then

#lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))#

(#oo/oo# and #0/0# are both indeterminate forms)

Essentially, whichever one gets where it's going faster (i.e. has a greater rate of change) wins.

In our case, the numerator is trying to go to zero and the denominator is trying to go to infinity, so if the numerator is growing faster we'll get #0# as our limit and if the denominator is growing faster we'll get #oo# as our limit.

Now onto the math:

We've established that plugging in zeros to #lim_(x->0) (5x^2)/(2x)# gives us an indeterminate form, so take

#f(x) = 5x^2# and #g(x) = 2x#

#f'(x) = 10x# and #g'(x) = 2# , so

#(f'(x))/(g'(x)) = (10x)/2 = 5x#

which means, by L'Hôpital's Rule:

#lim_(x->0) (5x^2)/(2x) = lim_(x->0) 5x = 0#

note: There are also many cases in which you do L'Hôpital's Rule and get another indeterminate form, in which case you have to apply L'Hôpital's Rule more than once.

Emilia Aguilar · Answer

undefined To evaluate the limit lim(5x^2)/(2x)dx as x approaches 0, we can simplify the expression by canceling out the common factor of x in the numerator and denominator. This gives us the limit lim(5x)/(2).

Now, substituting x=0 into the expression, we get (5*0)/(2) = 0/2 = 0.

Therefore, the limit of (5x^2)/(2x)dx as x approaches 0 is equal to 0.

How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x-&gt;0#?

How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x->0#?