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In case of $\delta$-$\epsilon$ definition of limit, books usually show that some $L$ is the limit of some function and then they prove it by reducing $L$ from the function $f(x)- L$ and showing $n>\epsilon$ and there is some $m\geq n$ ...and so on.
I also tried to calculate the limit of those functions by assuming $L$ to be a variable and equating the (example, $(n(1-L)+L)/\ldots$) $n = 0$ ...it gave me right value as limit without using L'Hopital rule.
To make my question more clear I am taking $f(n) = (2n+3)/(n-1)$ as an example.
Now, by the definition of limit it might have a limit, say $L$, and we need to find it. So, I use the definition of limit to find it in this way:
$$| (2n+3)/(n-1) - L)| < \epsilon \implies |(n(2-L)+(L+3))/(n-1)|$$
Equating product of $n$ I will get $L=2$ , which is the limit. In a way it says that the variable of the function has to go.
The question is: how shall it be done for quadratic and higher degree equations where numerator is of higher degree than denominator and trigonometric equations?
Elaborated examples,If $f(x)$= $|\frac{2n+3}{n-1}$-$L$|<$\epsilon$ $=>$ $|\frac{n(2-L)+(L+3)}{(n-1)}|$<$\epsilon$.
Now since we have decremented the limit from the $f(x)$, we expect it to be almost equal to $0$, so
$|\frac{n(2-L)+(L+3)}{(n-1)}|$=$0$ <br>
Hence $n$=$|\frac{(L+3)}{(L-2)}|$, Here L is the limit we expect to check and we can replace $n$ by 1,2,3...$N$ and it will give right result. <br>
In case of $n$=1, as the limit shall not exist, we get absurd result as <br>
$(3+$L$)=($L$-2)$
In case of $n$=+$\infty$, $($L$-2)=\frac{L+3}{\infty}$ which approaches to $0$,
hence $L$=2.
This works fine in case of higher degree equations also. But fails in case of equations that are indeterminate in nature.
**Note:Please try this once on your notebook.**
In case of $\delta$-$\epsilon$ definition of limit, books usually show that some $L$ is the limit of some function and then they prove it by reducing $L$ from the function $f(x)- L$ and showing $n>\epsilon$ and there is some $m\geq n$ ...and so on.
I also tried to calculate the limit of those functions by assuming $L$ to be a variable and equating the (example, $(n(1-L)+L)/\ldots$) $n = 0$ ...it gave me right value as limit without using L'Hopital rule.
To make my question more clear I am taking $f(n) = (2n+3)/(n-1)$ as an example.
Now, by the definition of limit it might have a limit, say $L$, and we need to find it. So, I use the definition of limit to find it in this way:
$$| (2n+3)/(n-1) - L)| < \epsilon \implies |(n(2-L)+(L+3))/(n-1)|$$
Equating product of $n$ I will get $L=2$ , which is the limit. In a way it says that the variable of the function has to go.
The question is: how shall it be done for quadratic and higher degree equations where numerator is of higher degree than denominator and trigonometric equations?
Elaborated examples,If $f(x)$= $|\frac{2n+3}{n-1}$-$L$|<$\epsilon$ $=>$ $|\frac{n(2-L)+(L+3)}{(n-1)}|$<$\epsilon$.
Now since we have decremented the limit from the $f(x)$, we expect it to be almost equal to $0$, so
$|\frac{n(2-L)+(L+3)}{(n-1)}|$=$0$ <br>
Hence $n$=$|\frac{(L+3)}{(L-2)}|$, Here L is the limit we expect to check and we can replace $n$ by 1,2,3...$N$ and it will give right result. <br>
In case of $n$=1, as the limit shall not exist, we get absurd result as <br>
$(3+$L$)=($L$-2)$
In case of $n$=+$\infty$, $($L$-2)=\frac{L+3}{\infty}$ which approaches to $0$,
hence $L$=2.
This works fine in case of higher degree equations also. But fails in case of equations that are indeterminate in nature.
**Note:Please try this once on your notebook.**
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