MHB Solving Limit for Horizontal Asymtotes

[tmt1]

We had to solve this limit

$$\lim_{{x}\to{\infty}} \frac{1}{x}$$

the answer is y= 0 is the Horizontal Asymptotes. I get the y = 0 but how do we know that it is the horizontal asymptote?

 

[Jameson]

I don't know how rigorous of an argument you are looking for but it's important to note that horizontal asymptotes only occur when we are looking at limits as $x \rightarrow \infty$ or $x \rightarrow -\infty$. As you know they are a value that a graph is forever leaning towards but never really reaches. We can't directly solve for this limit like we can for many other ones, but we see the value it is tending towards.

The limit you posted is a very standard starting point for looking at horizontal asymptotes. You will see that there are some rules for calculating them that have to do with the degree of the numerator compared to the degree of the denominator.

Just to show you another example, if you look at:

$\displaystyle \lim_{{x}\to{\infty}} \frac{x}{x^2+1}$

the answer is also 0 since the denominator grows faster than the numerator. Anyway, can you tell me what kind of insight you are looking for specifically for the problem you gave and maybe I can better explain it? :)

 

[tmt1]

I don't know how rigorous of an argument you are looking for but it's important to note that horizontal asymptotes only occur when we are looking at limits as $x \rightarrow \infty$ or $x \rightarrow -\infty$. As you know they are a value that a graph is forever leaning towards but never really reaches. We can't directly solve for this limit like we can for many other ones, but we see the value it is tending towards.

The limit you posted is a very standard starting point for looking at horizontal asymptotes. You will see that there are some rules for calculating them that have to do with the degree of the numerator compared to the degree of the denominator.

Just to show you another example, if you look at:

$\displaystyle \lim_{{x}\to{\infty}} \frac{x}{x^2+1}$

the answer is also 0 since the denominator grows faster than the numerator. Anyway, can you tell me what kind of insight you are looking for specifically for the problem you gave and maybe I can better explain it? :)

Thanks I think that is what I needed to know, I was just confused by some of my notes from the lecture. How can you tell if a limit has Vertical Asymptotes?

 

[Jameson]

Vertical asymptotes occur when you divide by 0 essentially. So if we use the example of: $$f(x)=\frac{1}{x}$$ there is one vertical asymptote at $x=\text{some value}$. Care to guess what that value is?

 

[tmt1]

Vertical asymptotes occur when you divide by 0 essentially. So if we use the example of: $$f(x)=frac{1}{x}$$ there is one vertical asymptote at $x=\text{some value}$. Care to guess what that value is?

When x = 0. Thanks, so if there is a vertical asymptote for a function, it occurs when the denominator is equal to 0?