When looking at a fraction of a limit to infinity, you need to only look at the highest power terms of the denominator and numerator.
I'm in a beginning calculus class after not touching math of any sort for 4 years. I've forgotten a lot of the more complicated algebra, and I'm struggling with simplifying limits right now. I'm doing my homework right now and coming across a lot of problems I'm getting wrong and I can't understand why, so I am asking for help.
To work this one out, I multiplied the top and bottom by ,
so I get
From there, I divided everything by since that's the highest x value in the denominator in order to get rid of some of the numbers. I wind up with:
Which is which equals
The back of my book however says this answer should just be 3. Why is the x^3 cancelled?
Here's another I got wrong:
All I could think to do was to divide everything by X but I don't even think you're allowed to do that without a denominator (are you?). Anyway when I do that I get
Apparently the answer is supposed to be 1/6, how do you get this?
The first is already in form, and the second can be made into that form using this transformation...
Recall that because of the continuity of the logarithm, , and the "stuff" inside the logarithm is now of the form . So you can use L'Hospital's Rule there too
Thanks for the first reply, I realized i made a stupid mistake on the first problem i posted and after working that out I got 3. Conjugating first was all I needed for the second problem as well.
Another question, How do you find horizontal and vertical asymptotes? I know for horizontal you have to make the limit of a function + or - infinity, and I'd assume for vertical it's the limit of a function when x approaches 0? I can provide a practice problem if it would be easier to grasp what im saying.
Horizontal--you need to check the limits at negative and positive infinity.
Slant--when the highest degree of the numerator is greater than the highest degree of the denominator. To solve, you do long division and disregard the remainder.