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Homework Statement
Hello! I got few questions, about limits.
º [tex]\lim_{x \rightarrow 0}(\frac{sin(x)}{x})=1[/tex]
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 [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?
Homework Equations
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}
xn>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.