$x_0=1$ $$\lim_{x\to x_0,\;x \,< \,x_0}\frac{f(x)-f(x_0)}{x-x_0}$$ where $f:R\to R,\;$ and $$f(x)=\left | x-1 \right |+\left | x+1 \right |$$ I calculate it and I get $\lim_{x\to 1,\;x \,< \,1} \frac{2-2}{x-1}$ which is $\frac{0}{0}$. I do not know what to say after this. Maybe I did something wrong but I thought that maybe that $\frac{0}{0}=0$ cause you have 0 divided by something almost $0$.
What is the following limit $\lim_{x\to x_0,\;x \,< \,x_0}\frac{f(x)-f(x_0)}{x-x_0}$, given $f(x)$, $x_0$
1
$\begingroup$
calculus
limits
-
1Hi! You're not new on the site -- would you mind sharing your own thoughts on this question? – 2017-02-15
-
1Here is a hint. Say $x_0-x = 1-x = h$ – 2017-02-15
-
1@amWhy thanks for the edit – 2017-02-15
-
0@DougM you do not get the same thing? – 2017-02-15
2 Answers
2
This is good. $\lim_\limits {x\to 1^-} \frac {2-2}{x-1}$
However this does not mean that $\frac 00 = 0$ It most certainly does not. $\frac 00$ is "indeterminate" It can really be anything.
However you can say: $\lim_\limits {x\to 1^-} \frac {0}{x-1} = 0$
When $x$ is in the neighborhood of $1$ (and less than $1$) the fraction equals 0.
-
0One could also say that the limit is the (one-sided) derivative of a constant function. – 2017-02-15
-
0it is constant in the interval (-1,1)... – 2017-02-15
0
Take a value a little bit less than 1 for x, say 0.9, and put it in for x, now evaluate. Then plug in 0.99, then 0.999, and so on. You should see the answer approaching a certain number. This is the limit as x approaches 1 from below.