lim x=a x^5-a^5/x^2-a^2

Question

question is- lim_x=a x⁵-a^5⁄x²-a²

Answer 1

You can also apply L'Hopital's rule, since the form is indeterminate.  You can apply it more than once until you get a determinate form.  By the way, you should group things properly in parentheses:

 

(x⁵ - a⁵)/(x² - a²)  

 

I assume that is what you really meant.  But for the expression as LITERALLY stated, then you get the result in Alexander's answer.

Answer 2

The notation is not exactly clear, but based on the qualified guess that the expression is actually x⁵ - a⁵/x² - a² it should be: a⁵ - a³ - a²

Answer 3

There is a trick here!

You will find that (x-a) is a factor of both the numerator and denominator.

Since for purposes of the limit x ≠ a, you can divide out (cancel) the factor x-a.

Then the limit will become obvious, but I will not do it for you...the divisions are good practice in algebraic manipulation.