## mytyl one year ago In session 5, clip 5, if X never equals to X(zero), how can X-X(zero)=0?

1. phi

$x - x_0 \ne 0$ rather he is saying $\lim_{x \rightarrow x_0} x-x_0 =0$ He could have been a bit clearer by writing $\lim_{x \rightarrow x_0} \frac{f(x)}{x-x_0 }(x-x_0)$ and then using the theorem that the limit of a product is the product of the limits, to say $\lim_{x \rightarrow x_0} \frac{f(x)}{x-x_0 } \cdot \lim_{x \rightarrow x_0}(x-x_0)$ and then $f'(x)\cdot 0 = 0$ For more backround see http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-4-derivatives-and-limits/ and the following lecture http://ocw.mit.edu/resources/res-18-006-calculus-revisited-single-variable-calculus-fall-2010/part-i-sets-functions-and-limits/lecture-5-a-more-rigorous-approach-to-limits/

2. mytyl

So......$\lim_{x \rightarrow x_{0}}(x-x_{0})=0?$

3. phi

yes

4. anonymous

mytyl is right to be confused by this. In the lecture, the professor makes an off the cuff suggestion that the second part of the above limit is solved by making $x =x_{o}$ He then, later in the lecture, takes specific time out of the lecture to state that it's important to remember that when solving the above limit, $x \neq x{o}$ It's still correct to say that $\lim_{x \rightarrow x{o}} (x - x{o}) = 0$ but the way the professor explained it was a bit confusing.