I will answer #1. repost for #2 and #3 if necessary.
A vertical asymptote is typically caused by the denominator becoming zero. The function is undefined at this point. The limits of the function from left and right may or may not be the same, and are usually ±&infin.
A removal discontinuity is apparent when you have a common factor in the numerator and the denominator, which therefore "cancel". This has the effect that the limits from the left and right of the discontinuity exist and are equal, but the function does not exist at the point of discontinuity.
In the example above, there is one of each.
i needed help on some problems and clarification on one.
1. Determine the location and type (removable, jump, infinite, or other) of all discontinuities of the function
(x^2-3x+2)/(x^2-1)
this section has been confusing for me for some reason, can any one briefly explain the rules and differences of discontinuities here, like how can one tell that the problem above is going to be a jump or removable or infinite discontinuity with out graphing. Also, what are there differences on a graph.
2. Lim as x->9 (2-sqrt(x-5))/(x-9)
please show me how to do this with out using any future methods such as l'hospitals rule. Right now it's in the 0/0 form and i am having trouble simplifying it. I tried the conjugates but still ending up with 0/0
3. Use the intermediate value theorem in order to show that the equation x^5-x+1=0 has at least one real solution
for this one i got "There is a zero some where between [-1,-2] because the sign changes some where in-between".
1 answer
2 answers