# On the algebra of Limits

• Likith D
Then you will have some power of ##x## in the denominator, which will cancel one of the powers in the numerator. You can also expand ##\sin(x)## and ##\cos(x)## to higher orders, but this is a PITA. In the end you should get terms that don't depend on ##x## at all, and you can just take the limit.You are correct that the terms in ##\frac{1}{\tan^2(x)}## will be messy. You can deal with this by writing ##\tan(x) = x + O(x^3)##. Then ##\tan^2(x) = x^2 + O(x

## Homework Statement

Find y;
$$y=\lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}$$

## Homework Equations

$$\lim_{x \rightarrow 0} {\frac{tan(x)}x}=1$$
$$\lim_{x \rightarrow 0} {\frac{sin(x)}x}=1$$

## The Attempt at a Solution

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}}\\
& = 0\\
\end{align}

or

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-cot^2(x)}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}+1-\frac {1} {sin^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{sin^2(x)}+1}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}+1}\\
& = 1\\
\end{align}

Plot on a graph, the solution seems to be $$\frac 2 3$$, which i do know how to come to using simplification and L'Hospital rule
But, I would like to know where I went wrong with my steps above

First of all: You are missing several parentheses that you really should have included.

It is not true that ##\lim_{x\to 0}[f(x) - g(x)] = \lim_{x\to 0}f(x) - \lim_{x\to 0}g(x)##. This is only true when all of these limits are well defined, which is not the case for the limits of 1/x^2 and 1/tan(x)^2. You need to actually manipulate the expression inside the parenthesis to produce an expression that converges properly.

Likith D
In the step from 1 to 2
and from 6 to 7

You can't write something like ##\infty - \infty = 0 ##

Orodruin said:
First of all: You are missing several parentheses that you really should have included.

It is not true that ##\lim_{x\to 0}[f(x) - g(x)] = \lim_{x\to 0}f(x) - \lim_{x\to 0}g(x)##. This is only true when all of these limits are well defined, which is not the case for the limits of 1/x^2 and 1/tan(x)^2. You need to actually manipulate the expression inside the parenthesis to produce an expression that converges properly.
But, it all happens such that the sum rule for derivatives of that function is true
...even when limits don't satisfy a sum rule?

Likith D said:
But, it all happens such that the sum rule for derivatives of that function is true

The sum rule for derivatives is only true when the derivative's involved exist. Don't neglect the words involved in the rules for derivatives. The rules are not merely equations.

For example, let ##f(x) = |x|## , ##g(x) = -|x|## and ##h(x) = f(x) +g(x)##. The derivative of ##h(x)## exists at ##x = 0##, but the derivatives of ##f(x)## and ##g(x)## do not.

Likith D said:

## Homework Statement

Find y;
$$y=\lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}$$

## Homework Equations

$$\lim_{x \rightarrow 0} {\frac{tan(x)}x}=1$$
$$\lim_{x \rightarrow 0} {\frac{sin(x)}x}=1$$

## The Attempt at a Solution

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}}\\
& = 0\\
\end{align}

or

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-cot^2(x)}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}+1-\frac {1} {sin^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{sin^2(x)}+1}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}+1}\\
& = 1\\
\end{align}

Plot on a graph, the solution seems to be $$\frac 2 3$$, which i do know how to come to using simplification and L'Hospital rule
But, I would like to know where I went wrong with my steps above

You seem to have had some vague idea of something that might work,But what you write is not very logical. Your (2) just repeats (1). Then you're taking to the limits looks based upon something that works for the reciprocals of those quantities. The ones of the problem are Infinite and, as has been said, the difference between two infinite quantities might be zero, finite or infinite.

Questions about limits are among the most common, possibly the most common, ones here. This one is more difficult than average.That suggests you have had problems or been given examples before, that would be helpful for you to look back on.

Otherwise it looks to me you are going to have to subtract an algebraic function from a trigonmetrical one. Then thinking logically you might think there is no way to do that, except if you can make the trigonometric one look like an algebraic one, i.e. expand It as a McLaurin series. This is not very easy for this function, but hopefully you only need a couple of terms, though you will need also to justify that.

Likith D said:

## Homework Statement

Find y;
$$y=\lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}$$
You mean ##y = \lim_{x \rightarrow 0} \{ \frac{1}{x^2} - \frac{1}{tan^2(x)}\}## In latex brackets"{" don't show up as brackets. If you want to print a bracket, put a backslash in front of it like "\{".

how to come to using simplification and L'Hospital rule

##\lim_{x \rightarrow 0} \{ \frac{1}{x^2} - \frac{1}{tan^2(x)}\} = \lim_{x \rightarrow 0} \{ \frac{1}{x^2} - \frac{\cos^2(x)}{sin^2(x)}\}##
## = \lim_{x \rightarrow 0} \frac{sin^2(x) - x^2 cos^2(x)}{x^2 sin^2(x)}##

I haven't worked this myself, but you might make progress by applying l'hospital's rule several times. Eventually you will get terms with no ##x## in them.

.

I don't think it is necessary to use l'Hopital. I think the fastest way is to expand tan(x) keeping more than the leading term.

Orodruin said:
I don't think it is necessary to use l'Hopital. I think the fastest way is to expand tan(x) keeping more than the leading term.
I think you are talking about taylor expansion of tan(x) term, but I'm unable to get simplifications for taylor expansion of tan(x) raised to power of -2. So, I'm not able to proceed further. Did you mean the taylor expansion of $$\frac 1 {tan^2(x)}$$?

Likith D said:
I think you are talking about taylor expansion of tan(x) term, but I'm unable to get simplifications for taylor expansion of tan(x) raised to power of -2. So, I'm not able to proceed further. Did you mean the taylor expansion of $$\frac 1 {tan^2(x)}$$?
Start by expanding ##\tan(x)## to order ##x^3##.

Orodruin said:
Start by expanding ##\tan(x)## to order ##x^3##.
oh, yeah, I got it, thanks!

Likith D said:
oh, yeah, I got it, thanks!
Do you mean that you got that particular part or that you managed to get the correct limit? It is difficult to tell if you don't write it out explicitly.

Orodruin said:
Do you mean that you got that particular part or that you managed to get the correct limit? It is difficult to tell if you don't write it out explicitly.
Well, yes, I did simplify the equation to get the answer as 2/3
And yes I find this method much easier compared to l'hospital's rule and repeated simplification , thanks for letting me know!

Key to your success was doing the check that showed that your first answer must be wrong.

Likith D said:
Well, yes, I did simplify the equation to get the answer as 2/3
And yes I find this method much easier compared to l'hospital's rule and repeated simplification , thanks for letting me know!
Don’t get too attached, in order to do the Taylor expansions you actually have to do the derivatives so it is effectively very similar to using l’Hopital ... It is just much easier to do it this way if you already are familiar with the expansions of some common functions.

Likith D
Likith D said:

## Homework Statement

Find y;
$$y=\lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}$$

## Homework Equations

$$\lim_{x \rightarrow 0} {\frac{tan(x)}x}=1$$
$$\lim_{x \rightarrow 0} {\frac{sin(x)}x}=1$$

## The Attempt at a Solution

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{tan^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}}\\
& = 0\\
\end{align}

or

\begin{align}
y & = \lim_{x \rightarrow 0} {\frac {1} {x^2}-cot^2(x)}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}+1-\frac {1} {sin^2(x)}}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{sin^2(x)}+1}\\
& = \lim_{x \rightarrow 0} {\frac {1} {x^2}}-\lim_{x \rightarrow 0}{\frac{1}{x^2}+1}\\
& = 1\\
\end{align}

Plot on a graph, the solution seems to be $$\frac 2 3$$, which i do know how to come to using simplification and L'Hospital rule
But, I would like to know where I went wrong with my steps above

If you insist on using l'Hospital's rule, re-write the problem first:
$$\frac{1}{x^2} - \frac{1}{\tan^2(x)} = \frac{\sin^2(x) - x^2 \cos^2(x)}{ x^2 \sin^2(x) }$$
Now that is in a form allowing straightforward application of l'Hospital, but an even faster way would be to expand the numerator and denominator in powers of ##x## and keep the lowest-order terms.

BTW: try to avoid writing ##tan(x)##, ##sin(x), ## or ##cos(x)##, etc; to get proper LaTeX (the way it was designed to be used) write instead ##\tan(x)##, ##\sin(x), ## and ##\cos(x)##; which you do by typing \tan instead of tan, \sin and \cos instead of sin and cos.

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