lim x--->1 x^2+2x-3/x-1
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If you really mean what you've written and the function whose limit you are taking is
x^2 + 2x - 3/x - 1
you have a rational function and are taking a limit at a point of its domain, so you can just plug the 1 in for x (since rational functions are continuous on their domain) and get 1 + 2 - 3 - 1 = -1
Unfortunately, I strongly suspect you meant to write
lim x--->1 (x^2 + 2x -3)/(x -1)
Again a rational function, but 1 is not in the domain since the denominator is 0 at x = 1. Fortunately, the numerator is also 0 at x = 1. So the trick is to factor the numerator, cancel common factors, giving a rational function which is equal to the original one, except at x = 1, then use the same reasoning as worked on the problem you probably typed by mistake.
So it becomes lim x--->1 (x-1)(x+3)/(x-1) = lim x--->1 (x+3) = 4.
(The first equation is true because a limit by definition only depends on what happens near the point being approached, in your problem 1. The second equation is true because for a rational function or polynomial, or more generally a continuous function, what happens to the value of the function as x approaches a point in the domain of the function, is it approaches the value of the function at the point.)
x^2 + 2x - 3/x - 1
you have a rational function and are taking a limit at a point of its domain, so you can just plug the 1 in for x (since rational functions are continuous on their domain) and get 1 + 2 - 3 - 1 = -1
Unfortunately, I strongly suspect you meant to write
lim x--->1 (x^2 + 2x -3)/(x -1)
Again a rational function, but 1 is not in the domain since the denominator is 0 at x = 1. Fortunately, the numerator is also 0 at x = 1. So the trick is to factor the numerator, cancel common factors, giving a rational function which is equal to the original one, except at x = 1, then use the same reasoning as worked on the problem you probably typed by mistake.
So it becomes lim x--->1 (x-1)(x+3)/(x-1) = lim x--->1 (x+3) = 4.
(The first equation is true because a limit by definition only depends on what happens near the point being approached, in your problem 1. The second equation is true because for a rational function or polynomial, or more generally a continuous function, what happens to the value of the function as x approaches a point in the domain of the function, is it approaches the value of the function at the point.)
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Is this
Lim (x² + 2x – 3)/(x – 1)?
x→1
If so, x² + 2x – 3 = (x – 1)(x + 3) so (x² + 2x – 3)/(x – 1) = (x + 3)/1
Therefore,
Lim (x²+ 2x – 3)/(x – 1) =
x→1
Lim (x + 3) = 4
x→1
You could apply l’Hopital’s rule, but the result is the same.
Lim (x² + 2x – 3)/(x – 1)?
x→1
If so, x² + 2x – 3 = (x – 1)(x + 3) so (x² + 2x – 3)/(x – 1) = (x + 3)/1
Therefore,
Lim (x²+ 2x – 3)/(x – 1) =
x→1
Lim (x + 3) = 4
x→1
You could apply l’Hopital’s rule, but the result is the same.
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x² + 2x - 3 simplifies to....
(x+3)(x-1)
Knowing that....
lim x---> 1: (x+3)(x-1) / (x-1)
Like terms cancel...
lim x---> 1: x-3
Plug in 1 for x...
Final Answer:
4
(x+3)(x-1)
Knowing that....
lim x---> 1: (x+3)(x-1) / (x-1)
Like terms cancel...
lim x---> 1: x-3
Plug in 1 for x...
Final Answer:
4