Given #(x−4)/ (x^2+6x−40)# how do you find the limit as x approaches 4?

Question

Christopher Allers · Accepted Answer

# 1/14#

well, we can plug in #x = 4# to see if there is a problem in the first place

for #f(x) = (x−4)/ (x^2+6x−40) #

#f(4) = (4−4)/ (16+24−40) = 0/0#

This is #0/0# indeterminate, so we are free to use L'Hôpital's Rule....

.....which tells us that #lim_(x to 4) (x−4)/ (x^2+6x−40)#

#= lim_(x to 4) (1)/ (2x+6)#

#= (1)/ (8+6) = 1/14#

But we might also have spotted that:

#f(x) = (x−4)/ (x^2+6x−40) = color(red)( (x−4)/ ((x-4)(x + 10))) = (x-4)/(x-4) * 1/(x + 10)#

And

#lim_(x to 4) (x-4)/(x-4) * 1/(x + 10)#

...using the idea that the limit of the product is the product of the limits #=lim_(x to 4) (x-4)/(x-4) * lim_(x to 4) 1/(x + 10) #

#= 1 * 1/14#

I was slightly surprised not to find a two-sided limit here, as they often come out of these kinds of question. But the fact that L'Hôpital's Rule led directly to the solution, as opposed to being one step off the two-sided solution, was the spur to factorise.

Axel Alvarez · Answer

undefined To find the limit as x approaches 4 for the expression (x−4)/ (x^2+6x−40), we can substitute the value of 4 into the expression and simplify. By substituting 4 for x, we get (4−4)/ (4^2+6(4)−40), which simplifies to 0/0. This is an indeterminate form. To evaluate the limit, we can factor the denominator and cancel out the common factor of (x−4). Factoring the denominator gives us (x−4)(x+10), and canceling out (x−4) from the numerator and denominator leaves us with 1/(x+10). Now, we can substitute 4 into the simplified expression, which gives us 1/(4+10) = 1/14. Therefore, the limit as x approaches 4 for the given expression is 1/14.