What is #lim_(x->5) (x^2-25)/(x-5)# ?

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Answer 1 · 2018-07-18T17:25:22

#10#

#\lim_{x\to 5}\frac{x^2-25}{x-5}#

#=\lim_{x\to 5}\frac{(x+5)(x-5)}{x-5}#

#=\lim_{x\to 5}(x+5)#

#=5+5#

#=10#

Answer 2 · 2018-07-18T20:00:02

#10#

#lim_(xto5)(x^2-25)/(x-5)#

#=lim_(xto5)(cancel((x-5))(x+5))/cancel((x-5))#

#=lim_(xto5)x+5=5+5=10#

Answer 3 · 2018-07-18T22:24:32

#10#

There are a few ways we can tackle this- I will show an algebraic way and a more calculus-themed way:

Algebraic Way:

If we plug #5# into our expression, we immediately see that we will get indeterminate form, or #0/0#.

The key is to realize that we have a difference of squares in the numerator. This allows us to rewrite the expression as

#((x+5)(x-5))/(x-5)#

Common factors cancel, and we're left with

#lim_(xto5) x+5=5+5=color(blue)(10)#

Calculus Way:

We can use L'Hôpital's rule, which says if we want to evaluate the limit of #f(x)/g(x)#, this equal to

#lim_(xtoc) (f'(x))/(g'(x))#

We take the derivative of our numerator and denominator expression, and evaluate the limit.

In our example,

#f(x)=x^2-25=>f'(x)=2x#

#g(x)=x-5=>g'(x)=1#

We now have the expression

#lim_(xto5)=(2x)/1=2*5=color(blue)(10)#

Both ways, we get #10#.

Hope this helps!