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lim (x^3-1)/(x-1), x-->1
First we try to solve the function by substituting x with 1.
then, lim (1^3 - 1)/(1-1)= 0/0
0/0 means that this method has failed because 1 is a root for both upper and lower functions,
Now we will try to factorize the function to eliminate the common factor
lim (x^3 - 1)/(x-1) = lim (x-1)(x^2+x+1)/(x-`1)
Now we substitute with 1
==> lim (x^2 +x+1) , x-->1 = lim(1^1+1+1) = 3
==> lim (x^3-1)/(x-1) = 3 when x-->1
To find the lt (x^3-1)/(x-1) as x-->1.
Let f(x) = (x^3-1)/(x-1). we see that at x=1 , f(1) = (1^1-1)/(1-1) takes the 0/0 form which is indeterminate.
In such cases we try to factorise numerator and denominator and reduce the rational function by the common factor and then take the limits. Numerator is x^3-1 = (x^2+x+1) ,
So, f(x) = (x-1)(x^2+x+1)/(x-1) = x^2+x+1. Therefore,
ltf(x) =[ Lt(x^2+x+1) as x--> 1 ] = 1^2+1+1+= 3.
In order to evaluate the limit, we'll choose the dividing out technique.
We'll apply the direct substitution, by substituting the unknown x, by the value1 and we'll see that it fails, because both, numerator and denominator, are cancelling for x=1.
Also, because x=1 is a root for both, that means that (x-1) is a common facor for both.
We'll write the numerator using the formula:
Now, we'll evaluate the limit:
lim (x^3-1)/(x-1) = lim (x-1)(x^2+x+1)/(x-1)
Now, we can divide out like factor:
lim (x^3-1)/(x-1) = lim (x^2+x+1)
We can apply the replacement theorem and we'll get:
lim (x^2+x+1) = 1^2 + 1 + 1 = 3
So, lim (x^3-1)/(x-1) = 3.
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