Understanding the Importance of Limits in Calculus

[kenewbie]

$$\lim_{x\to 2}\frac{3x^2-12}{x-2} = \lim_{x\to 2}\frac{3(x-2)(x+2)}{x-2} = \lim_{x\to 2}\frac{3(x+2)}{1} = 3*(2+2) = 12$$

This gives the exact same result as reducing the expression would do.

$$\frac{3x^2-12}{x-2} = 3(x+2)$$

I can see nothing but benefits by doing this, I get the same values for all X, and at x = 2 I no longer have a division by zero, rather I get 12 (which is the limit of the original expression at x = 2 anyway).

So what do I need the limits for? All the examples my book has of limits can be solved by simply reducing the expression to a simpler form instead.

Is my book just bad?

k

 

[LCKurtz]

Try sin(x) / x as x --> 0.

 

[Mark44]

$$\lim_{x\to 2}\frac{3x^2-12}{x-2} = \lim_{x\to 2}\frac{3(x-2)(x+2)}{x-2} = \lim_{x\to 2}\frac{3(x+2)}{1} = 3*(2+2) = 12$$

This gives the exact same result as reducing the expression would do.

$$\frac{3x^2-12}{x-2} = 3(x+2)$$

No it doesn't. (3x² - 12)/(x - 2) is not defined for x = 2, while 3(x + 2) is defined for all real numbers. What you have done to simplify the expression on the left is to factor out (x - 2)/(x - 2), which is 1 as long as x != 2. If x = 2, this expression is undefined. The whole point of limits is to be able to determine the value of some expression for values of x near some specific number.

I can see nothing but benefits by doing this, I get the same values for all X, and at x = 2 I no longer have a division by zero, rather I get 12 (which is the limit of the original expression at x = 2 anyway).

So what do I need the limits for? All the examples my book has of limits can be solved by simply reducing the expression to a simpler form instead.

Is my book just bad?

Probably not, but I don't know what book you're using. What they are doing is starting with simple examples of limits, and will probably go to more involved examples (such as lim (sin x)/x as LCKurtz mentioned) that are not amenable to such simple tricks.

k