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## some concrete limits

1. The problem statement, all variables and given/known data

Hello! I got few questions, about limits.
º $$\lim_{x \rightarrow 0}(\frac{sin(x)}{x})=1$$
If I take values for x close to zero I get:
f(x)=sinx/x
f(0.1)=0.017453283
f(0.01)=0.017453292
as I can see it is not even close to 1.
What is the problem? Where I am doing wrong?

º $$\lim_{x \rightarrow \infty}(\frac{1}{x})=0$$
Now, for all integers I agree that $$\lim_{x \rightarrow \infty}(\frac{1}{x})=0$$ (thanks to HallsofIvy for $\infty$), but what for 1/2, 1/3, 1/4 ?
$$\lim_{x \rightarrow \infty}(\frac{1}{x})=\frac{1}{1/4}=4$$ and not 0 ?

º $$\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})$$
How will I find the bound of the expression above?

2. Relevant equations

3. The attempt at a solution
$$\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})$$
I understand that x<0 (so the values for x are tending to 0 from the left side), and
$$\lim_{n \rightarrow \infty}(x_n)=0$$

For example, I know how to find the bound for:
$$\lim_{x \rightarrow 2^+}(\frac{x}{x-2})$$
D=R\{2}
xn>2
$$\lim_{n \rightarrow \infty}(x_n)=2$$
$$x_n-2>0$$
$$\lim_{n \rightarrow \infty}(x_n-2)=0$$
so that:
$$\lim_{x \rightarrow 2^+}(\frac{x}{x-2})=\lim_{n \rightarrow \infty}(\frac{x_n}{x_n-2})=\frac{2}{\lim_{n \rightarrow \infty}(2-x_n)}=+\infty$$

Thanks in advance.

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Mentor

Quote by Дьявол 					 				 				1. The problem statement, all variables and given/known data  Hello! I got few questions, about limits. º $$\lim_{x \rightarrow 0}(\frac{sin(x)}{x})=1$$ If I take values for x close to zero I get: f(x)=sinx/x f(0.1)=0.017453283 f(0.01)=0.017453292 as I can see it is not even close to 1. What is the problem? Where I am doing wrong?
Are you using a calculator to do these? If so, I think your calculator is in degree mode. It needs to be in radian mode. As the values of x get smaller, the value of your expression will get closer to 1.

Quote by Дьявол 					 				 				 º $$\lim_{x \rightarrow \infty}(\frac{1}{x})=0$$ Now, for all integers I agree that $$\lim_{x \rightarrow \infty}(\frac{1}{x})=0$$ (thanks to HallsofIvy for $\infty$), but what for 1/2, 1/3, 1/4 ?  $$\lim_{x \rightarrow \infty}(\frac{1}{x})=\frac{1}{1/4}=4$$ and not 0 ?
The limit is for x growing very large, so you shouldn't concern yourself with small values of x. On the other hand,
$$\lim_{x \rightarrow 0^+}(\frac{1}{x})=\infty$$
which is more related to what you're doing with 1/2, 1/3, and so on.

Quote by Дьявол 					 				 				$$\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})$$ How will I find the bound of the expression above?
As x approaches 0 from the negative side, 1/x approaches neg. infinity, so e^(1/x) approaches 0. Do you need more explanation than that?

Quote by Дьявол 					 				 				2. Relevant equations    3. The attempt at a solution $$\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})$$ I understand that x<0 (so the values for x are tending to 0 from the left side), and $$\lim_{n \rightarrow \infty}(x_n)=0$$  For example, I know how to find the bound for:  $$\lim_{x \rightarrow 2^+}(\frac{x}{x-2})$$ D=R\{2} xn>2 $$\lim_{n \rightarrow \infty}(x_n)=2$$ $$x_n-2>0$$ $$\lim_{n \rightarrow \infty}(x_n-2)=0$$ so that: $$\lim_{x \rightarrow 2^+}(\frac{x}{x-2})=\lim_{n \rightarrow \infty}(\frac{x_n}{x_n-2})=\frac{2}{\lim_{n \rightarrow \infty}(2-x_n)}=+\infty$$  Thanks in advance.

Blog Entries: 1
Thanks for the post Mark44.

Yes, I was using calculator in degree mode. Now with radian mode everything is all right.

For the second one. Sorry, I wasn't so clear. I was learning about the number "e". So for one task (example):
$$\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})=\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})^{(x-3)+3}=\lim_{t \rightarrow \infty}(1+\frac{1}{t})^t*\lim_{t \rightarrow \infty}(1+\frac{1}{t})^3=e*(1+0)^3=e$$
As we can see they put $$\lim_{t \rightarrow \infty}(\frac{1}{t})=0$$. How is this possible? What about for t=1/2,1/3,1/4 ? It wouldn't be zero in that case.

Quote by Mark44 					 				 				As x approaches 0 from the negative side, 1/x approaches neg. infinity, so e^(1/x) approaches 0. Do you need more explanation than that?
Mark44, sorry for misunderstanding again. Yes I understand all of that, but how will I "show" or "prove" that. Aren't there any calculations?
Thanks in advance.

Mentor

## some concrete limits

Quote by Дьявол 					 				 				Thanks for the post Mark44.  Yes, I was using calculator in degree mode. Now with radian mode everything is all right.   For the second one. Sorry, I wasn't so clear. I was learning about the number "e". So for one task (example): $$\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})=\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})^{(x-3)+3}=\lim_{t \rightarrow \infty}(1+\frac{1}{t})^t*\lim_{t \rightarrow \infty}(1+\frac{1}{t})^3=e*(1+0)^3=e$$
For your first limit above, I think you are missing an exponent of x. In other words, I think it should be:
$$\lim_{x \rightarrow \infty}(1+\frac{1}{x-3})^x$$
Now I think I understand what you're saying. The first limit was as x approached infinity and involved an expression with (x - 3). They substituted t = x - 3 and changed the limit variable from x to t (as x gets very large, so does t).

In the third limit expression (before they took the limit), there are two factors:
$$}(1+\frac{1}{t})^t$$ and
$$}(1+\frac{1}{t})^3$$.
The second one is straightforward to evaluate in the limit, and turns out to be just 1. If you multiply it out before taking the limit, you have 1 + 3*1/t + 3*1/t^2 + 1/t^3, which approaches 1 as t gets large.

The first one is more tricky, and you can't just say that 1/t approaches 0 as t gets large. There are two competing effects going on: the base, 1 + 1/t, is getting closer to 1, but the exponent t is getting larger. The net effect is that (1 + 1/t)^t approaches the number e as t gets large. If I recall correctly, one of the definitions of e is precisely this limit.

Quote by Дьявол 					 				 				As we can see they put $$\lim_{t \rightarrow \infty}(\frac{1}{t})=0$$. How is this possible? What about for t=1/2,1/3,1/4 ? It wouldn't be zero in that case.

Quote by Дьявол 					 				 				  Mark44, sorry for misunderstanding again. Yes I understand all of that, but how will I "show" or "prove" that. Aren't there any calculations? Thanks in advance.

Blog Entries: 1  Now I understand. I misjudged $\infty$.  But for $$\lim_{x \rightarrow 0^-}(e^{1/x})$$, I need to "show", "explain" the result of the limit. How will I do that?  Regards.

Recognitions:      Gold Member        Science  Advisor     Staff Emeritus  For x very close to 0, but negative, say -0.00001, 1/x is -100000, a very large negative number.  What is e-100000?  what is e any very large negative number?

Blog Entries: 1  Thanks for the post. It definitely will tend to zero.  And what about $$\lim_{x \rightarrow 0^+}(\frac{1}{1+e^{1/x}})$$?  $$\lim_{x \rightarrow 0^+}(e^{1/x})=\infty$$ ?

Mentor  For the first one, as x approaches 0 (from the right), 1/x grows without bound (approaches infinity), so 1 + 1/x also grows without bound, which makes the fraction approach 0.  For the second, that's the right value.

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