Evaluate the limit of the function (x^2-1)/(x-1) if x goes to 1?
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Since `(x^2-1) = (x-1)(x+1)`
`lim_(x->1) (x^2-1)/(x-1) = lim_(x->1) ((x+1)(x-1))/(x-1) = lim_(x->1) (x+1) = 2`
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The limit `lim_(x->1) (x^2-1)/(x-1)` has to be determined.
If we substitute x = 1 in the polynomial `(x^2-1)/(x-1)` we get `0/0` which is indeterminate.
One way around this is by factorizing the denominator and numerator and canceling common terms.
`lim_(x->1) (x^2-1)/(x-1)`
Use the property `x^2 - y^2 = (x - y)(x + y)`
`lim_(x->1) ((x - 1)(x + 1))/(x-1)`
= `lim_(x->1) (x + 1)`
Now substituting x = 1 gives 1 + 1 = 2
The limit `lim_(x->1) ((x - 1)(x + 1))/(x-1) = 2`
We'll substitute x by 1 in the expression of the function:
lim (x^2 - 1)/(x-1) = (1-1)/(1-1) = 0/0
x->1
Since we've get an indetermination, we can apply L'Hospital's rule:
lim f(x)/g(x) = lim f'(x)/g'(x)
Let f(x) = x^2 - 1 => f'(x) = 2x
Let g(x) = x - 1 => g'(x) = 1
lim (x^2 - 1)/(x-1) = lim 2x/1
x->1 x->1
We'll replace x by 1:
lim 2x/1 = 2/1 = 2
The requested limit of the given function, if x approaches to 1, is lim (x^2 - 1)/(x-1) = 2.
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