(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Hello! I got few questions, about limits.

º [tex]\lim_{x \rightarrow 0}(\frac{sin(x)}{x})=1[/tex]

If I take values forxclose 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 to1.

What is the problem? Where I am doing wrong?

º [tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=0[/tex]

Now, for all integers I agree that [tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=0[/tex] (thanks to HallsofIvy for [itex]\infty[/itex]), but what for 1/2, 1/3, 1/4 ?

[tex]\lim_{x \rightarrow \infty}(\frac{1}{x})=\frac{1}{1/4}=4[/tex] and not 0 ?

º [tex]\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})[/tex]

How will I find the bound of the expression above?

2. Relevant equations

3. The attempt at a solution

[tex]\lim_{x \rightarrow 0^-}(e^{\frac{1}{x}})[/tex]

I understand that x<0 (so the values for x are tending to 0 from the left side), and

[tex]\lim_{n \rightarrow \infty}(x_n)=0[/tex]

For example, I know how to find the bound for:

[tex]\lim_{x \rightarrow 2^+}(\frac{x}{x-2})[/tex]

D=R\{2}

x_{n}>2

[tex]\lim_{n \rightarrow \infty}(x_n)=2[/tex]

[tex]x_n-2>0[/tex]

[tex]\lim_{n \rightarrow \infty}(x_n-2)=0[/tex]

so that:

[tex]\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[/tex]

Thanks in advance.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Some concrete limits

**Physics Forums | Science Articles, Homework Help, Discussion**