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How do you find the limit of #(x^2-1)/(x-1)# as #x->1#?

Answer 1

# lim_(x rarr 1)(x^2-1)/(x-1) = 2 #

Let # f(x) = (x^2-1)/(x-1) # then #f(x)# is defined everywhere except at #x=1#, however when we evaluate the limit we are not interested in the value of #f(1)#, just the behaviour of #f(c)# for #c# close to #1#.

# lim_(x rarr 1)(x^2-1)/(x-1) = lim_(x rarr 1)((x+1)(x-1))/((x-1)) # # :. lim_(x rarr 1)(x^2-1)/(x-1) = lim_(x rarr 1) (x+1), " as " x!= 1# # :. lim_(x rarr 1)(x^2-1)/(x-1) = 1+1 # # :. lim_(x rarr 1)(x^2-1)/(x-1) = 2 #

graph{(x^2-1)/(x-1) [-10, 10, -5, 5]}

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Answer 2

Luke Alvarez

To find the limit of (x^2-1)/(x-1) as x approaches 1, we can simplify the expression by factoring the numerator. This gives us (x+1)(x-1)/(x-1). Since (x-1) appears in both the numerator and denominator, we can cancel it out. This leaves us with (x+1). Now, we can substitute x=1 into the simplified expression, which gives us (1+1) = 2. Therefore, the limit of (x^2-1)/(x-1) as x approaches 1 is equal to 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #(x^2-1)/(x-1)# as #x-&gt;1#?

How do you find the limit of #(x^2-1)/(x-1)# as #x->1#?