Evaluate the following

Question

evaluate the limit

Answer 1

We want to find

lim_{{x → -1}} (x⁶-1)/(x⁴-1)

First we try to plug in x=-1 into the expression to see what happens.

( (-1)⁶-1 ) / ( (-1)⁴-1 ) = 0/0

Since we get 0/0, this means we can try to simplify the expression, cancel some terms, and then try to plug in x=-1 again. We will factor the numerator and denominator using the difference of two squares

x⁶-1 = (x³-1)*(x³+1)

x⁴-1 = (x²-1)*(x²+1)

So the limit becomes

lim_{{x → -1}} (x⁶-1)/(x⁴-1)

= lim_{{x → -1}} ( (x³-1)*(x³+1) ) / ( (x²-1)*(x²+1) )

Next we use the difference of two cubes, the sum of two cubes, and difference of two squares formulas

x³-1 = (x-1)*(x²+x+1)

x³+1 = (x+1)*(x²-x+1)

x²-1 = (x-1)*(x+1)

So the limit becomes

lim_{{x → -1}} (x⁶-1)/(x⁴-1)

= lim_{{x → -1}} ( (x³-1)*(x³+1) ) / ( (x²-1)*(x²+1) )

= lim_{{x → -1}} ( (x-1)*(x²+x+1)*(x+1)*(x²-x+1) ) / ( (x-1)*(x+1)*(x²+1) )

Next we can cancel both the (x-1) and (x+1) terms in the numerator and denominator

lim_{{x → -1}} (x⁶-1)/(x⁴-1)

= lim_{{x → -1}} ( (x³-1)*(x³+1) ) / ( (x²-1)*(x²+1) )

= lim_{{x → -1}} ( (x-1)*(x²+x+1)*(x+1)*(x²-x+1) ) / ( (x-1)*(x+1)*(x²+1) )

= lim_{{x → -1}} ( (x²+x+1)*(x²-x+1) ) / (x²+1)

Finally, try to plug in x=-1 to this simplified expression to see if the limit exists

( ((-1)²+(-1)+1)*((-1)^2-(-1)+1) ) / ( (-1)² + 1) = ((1)*(3))/2 = 3/2

Since we got a number for the limit, then the limit must be equal to 3/2

Answer 2

Because evaluating the function at x = -1 yields the indeterminate form 0 / 0 , we can apply L'Hospital's rule by taking the derivatives of the numerator and denominator:

lim_x→-1 6x⁵ / 4x³ = 3/2