How do you graph a function such as $$f(x)=\frac{x^2+3x+2}{x+1}$$ and find its limits $\lim_{x\to-1^-}f(x)$, $\lim_{x\to-1^+}f(x)$, $\lim_{x\to-1}f(x)$?
Thank you!
How do you graph a function such as $$f(x)=\frac{x^2+3x+2}{x+1}$$ and find its limits $\lim_{x\to-1^-}f(x)$, $\lim_{x\to-1^+}f(x)$, $\lim_{x\to-1}f(x)$?
Thank you!
I personally find it easier to calculate limits, vertical asymptotes and several other things, before trying to actually draw a graph.
Here's what i'd suggest you to do
In the majority of cases (at least for simple functions) you can make a pretty good guess, what the graph will look like and you can sketch it along your calculated points and towards the asymptotes.
As you also asked how to calculate limits for functions in this fractional form, you could try to apply the limit in the numerator and denominator seperately (if they exist!!). However you have to be careful! Cases that result in something like "$\frac{0}{0}$" or "$\frac{\pm\infty}{\pm\infty}$" cannot be solved that way. In those cases you could try using l'Hopital's rule.
You have a function defined as the ratio of two polinomials. Since the degree of the numerator is not less than the degree of the denominator, you could make the polinomial division to simplify the expression. Eventually you find (as suggested in the comments) that the numerator is a multiple of the denominator (the rest is 0) so that the function $f$ can be written as: $$ f(x) = x+2 \qquad \text{with $x\neq -1$} $$ So the graph of the function is given by the line $y=x+2$ with the point $(-1,1)$ removed. The limits as $x\to -1$ of $f$ are equal to the limits as $x\to -1$ of $g(x) = x+2$, which is defined in $x=-1$ and is continuous. Hence those limits can be computed simply by computing $g(-1) = -1 + 2 = 1$. This is apparent looking at the graph, anyway.