# Finding Limit Function that satisfies the conditions.

I am having some trouble figuring out a few math problems from my Calc 1 class. I am not sure where to start, as all the limits are different.

1. find a function that satisfies the given conditions and then sketch it.

2. sketch a graph of the function y=f(x) that satisfies the given conditions. Just label the coordinate axes and sketch the appropriate graph.

For 60, 62, and 64 they are kinda the same thing. 60: would it be a function where if you let X=-1, and the denominator =0, is that what we are looking for? like $\frac{2}{x+1} \ .$

1. I would say yes, even though g(x) and f(x) are not discontinuous on their own, that changes when you put them in a function together such as $f(x)/g(x).$

2. not to sure

Thank you for all your help.

• For 60; would 2x/x+1 be an answer since it has a non removable discontinuity at x=-1 ? – user131785 Feb 27 '14 at 3:44
• Simple example for 62: $f(x) = 1$, $g(x) = x$. Or, if you want the discontinuity in the interior of $[0,1]$, you can use $g(x) = x - 1/2$. – Bungo Aug 22 '14 at 6:05

Just try drawing what the graph could potentially be on paper, and find an equation for it. Here's a freebie for the first one (number 74): $$g(x)=\dfrac{1}{x-3}$$ Note: I am not taking calculus yet, but I do know about limits and such... I don't know if there is a faster way because my way seems inefficient and not algebraic-like.
EDIT: For the second one, maybe this would work: $$f(x)=\begin{cases} -\frac{1}{x^2+\frac{1}{2}}, \ \ \ \ x < 0 \\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \\ \frac{1}{x^2+\frac{1}{2}}, \ \ \ \ \ \ \ \ x > 0 \\ \end{cases}$$ YET ANOTHER EDIT: Your answer for the final one (nonremovable discontinuity) is correct.
• @user131785 I think it is... There are many functions that are correct (e.g. $g(x)=\dfrac{x^2+2x+1}{x+1}$) – TrueDefault Feb 27 '14 at 3:52
• @user131785 To be totally honest with you I don't really know what a nonremovable discontinuity is. I think your answer $\dfrac{2x}{x-1}$ is better because it does not have any factors that can cancel out. I don't know if I am correct, but because my answer has factors that can cancel, it is removable. In any case you are right. – TrueDefault Feb 27 '14 at 3:58