# Need help with calculus hw please!

1. Dec 13, 2004

### yawie

Hi,
I have some questions that i need help on, they're due tomorrow, and i've been trying with out success the whole weekend!

a) lim e^(1/x) log(100+ x)
x->0-

b) lim ((x^2+3x-10)/(x-2)) log(10^2-x)
x->2

c) lim 6^x
x->-infinity

THanks alot in advance! I really appreciate it!
Yawie

2. Dec 13, 2004

### quasar987

1) If both the limit of e^(1/x) and of log(100 + x) separetely are real numbers, then you know the limit of the product is the product of those numbers. Now, what happens to 1/x as x approaches 0 from the left? You can easily convince yourself that this quotient decreases endlessly and very rapidly (either by ploting the graph or putting in numbers yourself). We therefor write that the limit as x approaches 0- is $-\infty$. So your limit is equivalent to

$$\lim_{y \rightarrow \infty} e^{-y} = \lim_{y \rightarrow \infty} \frac{1}{e^y}$$

, etc.

2) You would like to do the same thing as for #1; i.e. find the limit of the quotient of polynomial and the limit of the log separately, and multiply those limits together to get the result. But as you may have noticed, the limit of the polynomial is of the form 0/0. This ratio is undefined, i.e. it is not a real number. So you'll have to perform some algebraic manipulations on your quotient so it takes an acceptable form (note: whenever you have a limit of the kind of 0/0 or any of the other 6 indeterminate forms, there is always a way to transform it into something more docile.)
When you have to evaluate the limit of a quotient of polynomials, a trick that always work is to factor out an x to the largest power from the numerator and the denominator. Ex: in your case, the polynomial of higher degree is the one at the numerator; its degree is 2. All you have to do is factor x^2 from the num and denom, cancel them out. See what happens.

As for the limit of the log, remember that whatever^0 = 1 and log(1) = 0.

3) Same thing I said for a)

Last edited: Dec 13, 2004