What Are the Solutions to These Calculus Limits?

[yawie]

Hi,
I have some questions that i need help on, they're due tomorrow, and I've been trying without 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 a lot in advance! I really appreciate it!
Yawie

 

[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)

 

[wais]

Hi Yawie,

I am happy to help you with your calculus homework. It's great that you are reaching out for help instead of giving up. Let's take a look at the questions you have provided.

a) For this limit, we can use the L'Hopital's Rule. First, we will rewrite the expression as:
lim (e^(1/x) log(100+x)) = lim (log(100+x)/e^(-1/x))
Now, we can apply L'Hopital's Rule, which states that if we have an indeterminate form of 0/0 or infinity/infinity, we can take the derivative of the numerator and denominator separately to simplify the expression. In this case, we have 0/0, so we can apply the rule.
lim (log(100+x)/e^(-1/x)) = lim (((1/(100+x)))/(e^(-1/x) * (-1/x^2)))
= lim ((-x/(100+x)) * (e^(1/x)) * (1/x^2))
= lim ((-x/(100+x)) * (e^(1/x)) * (1/x^2))
Now, we can plug in x=0 to get the final answer:
lim (e^(1/x) log(100+x)) = (-0/100) * (1/1) = 0

b) For this limit, we can use the same approach as in part a). First, we will rewrite the expression as:
lim ((x^2+3x-10)/(x-2)) log(10^2-x) = lim ((x^2+3x-10)/(x-2)) * (log(10^2-x)/1)
Now, we can apply L'Hopital's Rule to the first part of the expression:
lim ((x^2+3x-10)/(x-2)) = lim ((2x+3)/(1)) = 5
For the second part, we can use a property of logarithms to rewrite it as:
log(10^2-x) = log(100) - log(10-x) = 2 - log(10-x)
Now, we can apply L'Hopital's Rule again:
lim (2 - log(10-x)) = lim (-1/(10-x)) = -1/8
Finally, we can multiply the two parts together to