Finding Tangent Slope at General Point with Limits

[Jenninifer]

I've been using the formula
lim
x-->a f(x)-f(a)
--------
x-a
I haven't had any problems until I was asked to find the slope of the tangent at the general point whose x-coordinate is a. How do I do it with an a instead of a point?

I'm trying to find it for x^2+x+1

Thanks!

 

[DonAntonio]

I've been using the formula
lim
x-->a f(x)-f(a)
--------
x-a
I haven't had any problems until I was asked to find the slope of the tangent at the general point whose x-coordinate is a. How do I do it with an a instead of a point?

I'm trying to find it for x^2+x+1

Thanks!

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

DonAntonio

 

[Jenninifer]

I got to the first step, but how do you get to the second step?

 

[Office_Shredder]

By the second step do you mean when he factors the numerator, or when he cancels the x-a from the numerator and the denominator?

 

[scurty]

I got to the first step, but how do you get to the second step?

$\displaystyle \lim_{x \rightarrow a} \frac{x^2+x+1-(a^2+a+1)}{x-a} = \lim_{x \rightarrow a}\frac{x^2+x-a^2-a}{x-a} = \lim_{x \rightarrow a}\frac{x^2+ax+x-a^2-ax-a}{x-a}$

Do you see what to do from there? Creatively adding 0 or multiplying by 1 shows up a lot in mathematics.

 

[DonAntonio]

$\displaystyle \lim_{x \rightarrow a} \frac{x^2+x+1-(a^2+a+1)}{x-a} = \lim_{x \rightarrow a}\frac{x^2+x-a^2-a}{x-a} = \lim_{x \rightarrow a}\frac{x^2+ax+x-a^2-ax-a}{x-a}$

Do you see what to do from there? Creatively adding 0 or multiplying by 1 shows up a lot in mathematics.

I think that when when we reach some level in mathematics, high school tricks only can, sometimes,

confuse and/or make things lengthier and more cumbersome.

Adding and substracting that "ax" in the numerator in one of those tricks, and in this case it just adds more writing innecessarily, imo.

The best, I think, is to observe that we have a difference of squares in the numerator and thus directly to factor:

$x^2+x-a^2-a=x^2-a^2+x-a=(x-a)(x+a)+x-a=(x-a)(x+a+1)$ ...

DonAntonio

 

[scurty]

Hmm, I didn't think of it that way! I like that way better!