Limits Help

1. (2 + e^1/x) / (2 - e^1/x)
What is the horizontal asymptotes?

2. f(x) = { sin x/x, x cant be 0
k, x = 0
In order for f(x) to be continous at x=0, the value of k must be ?

3. If a cant be 0, then Limit x->a (x^2 - a^2)/ (x^4 - a^4) is?

Help please? thankyou

Quote:

---

Originally Posted by tmd

(2 + e^1/x) / (2 - e^1/x)
What is the horizontal asymptotes?

---

I'm assuming this function is: (2 + e^{1/x})/(2 - e^{1/x})?
Horizontal asymptotes occur when we take the function limit at x -> (+/-) infinity.

So, as x -> - infinity, 1/x -> 0 so e^{1/x} -> 1 Thus your function approaches (2 + 1)/(2 - 1) = 2. Since this is a constant, we have a horizontal asymptote at y = 2 for x->-infinity.

As x -> infinity we have the same behavior for 1/x so the function again has a horizontal asymptote at y = 2 for x->infinity.

(Technically there are still two horizontal asymptotes, even though they are the same line.)

-Dan

Quote:

---

Originally Posted by tmd

f(x) = { sin x/x, x cant be 0
k, x = 0
In order for f(x) to be continous at x=0, the value of k must be ?

---

Are you saying that for x not equal to zero f(x) = sin(x)/x and at x = 0 f(x) = k?

For f(x) to be continous we must have that k = lim(x->0) sin(x)/x. There are various ways to find this limit. Do you know the power series expansion of sin(x)? If so, this is probably the simplest way to get this limit.

For x close to 0
sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...

So for x close to 0
sin(x)/x is approximately [x - (1/3!)x^3 + (1/5!)x^5 - ...]/x = 1 - (1/3!)x^2 + (1/5!)x^4 - ...
which is approximately 1. ( x^n << 1 for x close to 0 for all positive integers n)

Thus k = 1.

-Dan

Quote:

---

Originally Posted by tmd

If a cant be 0, then Limit x->a (x^2 - a^2)/ (x^4 - a^4) is?

---

x^4 - a^4 is the difference between two squares (x^2)^2 and (a^2)^2 so this factors as:
x^4 - a^4 = (x^2 + a^2)(x^2 - a^2)

so
(x^2 - a^2)/(x^4 - a^4) = 1/(x^2 + a^2)

So lim(x->a) (x^2 - a^2)/(x^4 - a^4) = 1/(a^2 + a^2) = 1/(2a^2)

-Dan

Quote:

---

Originally Posted by topsquark

For x close to 0
sin(x) = x - (1/3!)x^3 + (1/5!)x^5 - ...

---

No need to do that.
The limit of,
sin(x)/x
Is one of those students must know immediately.

Quote:

---

Originally Posted by tmd

1. (2 + e^1/x) / (2 - e^1/x)
What is the horizontal asymptotes?

---

Divide the numerator and demoninator by e^(1/x)
Thus,
(2e^(-1/x)+1)/(2e^(-1/x)-1)
When x---> +oo
The numerator is, 2(1)+1=3
The denominator is, 2(1)-1=1 not equal to zero.
Thus, by the quotient rule of limits this limit is, (3)/(1)=3
That is one horizontal asymptote.
---
Instead of dividing leave it the way it should be
(2 + e^(1/x)) / (2 - e^(1/x))
As x---> -oo
The numerator is 2+1=3
The denominator is 2-1=1 not equal to zero.
Thus, by the qtuotient rule of limits this limit is (3)/(1)=3
Also the same.