I have troubles finding the limit of this piecewise function

[El foolish Phenomeno]

Homework Statement

find the domain of the piece wise function , its limit at 2, and negative infinity , positive infinity and 1/2 and find a so that the function is continuous at 2

Relevant Equations

lim f(x) when x tends to y = L

I have troubles finding the limits at the designated points , should i only find the limit at infinity where f(x) has belongs to an interval containing inifinity? (sorry for english)

and for the a this is what i attempted. i am unsure.

Our textbook never talks about piecewise functions and their rules. If you havd any textbook recommandations or websites videos that can help me learn more abkut the calculus of piecewise functions. i will gladly take them

Thank you for your help and your time. It's greatly appreciated

 

[PeroK]

Our textbook never talks about piecewise functions and their rules. If you havd any textbook recommandations or websites videos that can help me learn more abkut the calculus of piecewise functions. i will gladly take them

There's no such thing as a "piecewise" function. It's just a regular function that is given by a different formula for different values of $x$. There are, therefore, no different rules.

The key point is that the limits, continuity and differentiability of these functions may not be obvious. In general, you have to check explicitly at the points where the function's formula changes. Have you studied one-sided limits? Like $\lim_{x \to 2^-}$ and $\lim_{x \to 2^+}$.

For the limits at $\pm \infty$ the usual definition applies. It should be clear that as $x \to \infty$ only one formula is relevant.

 

[El foolish Phenomeno]

There's no such thing as a "piecewise" function. It's just a regular function that is given by a different formula for different values of $x$. There are, therefore, no different rules.

The key point is that the limits, continuity and differentiability of these functions may not be obvious. In general, you have to check explicitly at the points where the function's formula changes. Have you studied one-sided limits? Like $\lim_{x \to 2^-}$ and $\lim_{x \to 2^+}$.

For the limits at $\pm \infty$ the usual definition applies. It should be clear that as $x \to \infty$ only one formula is relevant.

We have studied one-sided limits.

I don't know if i am getting you but from what i understand of your explanation "It should be clear that $x \to \infty$ only one formula is relevant." you meant that if x tends to +infinity i just need to study the formula defined in the interval [2, +infinity) , for negative infinity i should only use the the interval (-infinity, 1/2]. does the same logic apply to limits as tends to 2 and 1/2?

 

[Mark44]

We have studied one-sided limits.

I don't know if i am getting you but from what i understand of your explanation "It should be clear that t $x \to \infty$ only one formula is relevant." you meant that if x tends to +infinity i just need to study the formula defined in the interval [2, +infinity) , for negative infinity i should only use the the interval (-infinity, 1/2].

Yes to both questions.

does the same logic apply to limits as tends to 2 and 1/2?

Not exactly. Since 1/2 and 2 are endpoints of intervals in which different formulas apply to get the two-sided limits at each of these points you'll need to determine the left- and right-side limits at each of these points, meaning that you will need to use the appropriate function formula for a left-side limit or a right-side limit.

BTW, your posted pictures are unreadable, at least by me. Your handwriting is very difficult to read, and the extremely poor lighting in the images make things worse. For reasons of legibility we discourage the use of images that show the work done.

 

[PeroK]

We have studied one-sided limits.

Good. That's what you need if you have a different formula either side of a point. A limit exists iff the one-sided limits exist and are equal. That's a theorem!

I don't know if i am getting you but from what i understand of your explanation "It should be clear that $x \to \infty$ only one formula is relevant." you meant that if x tends to +infinity i just need to study the formula defined in the interval [2, +infinity) , for negative infinity i should only use the the interval (-infinity, 1/2]. does the same logic apply to limits as tends to 2 and 1/2?

Yes. You can prove these things if you know how. In general, if we are looking at the limit as $x \to x_0$, we can assume that $|x - x_0| < 1$, for example. That's sometimes useful.

Another example is where $x_0 > 0$, we can assume that $x > 0$ when looking at the limit. If you are using the epsilon-delta definition, then the justification is that we can take $\delta < x_0$. So that $|x - x_0| < \delta \ \Rightarrow \ x > 0$. Does that make sense?

 

[El foolish Phenomeno]

Good. That's what you need if you have a different formula either side of a point. A limit exists iff the one-sided limits exist and are equal. That's a theorem!

Yes. You can prove these things if you know how. In general, if we are looking at the limit as $x \to x_0$, we can assume that $|x - x_0| < 1$, for example. That's sometimes useful.

Another example is where $x_0 > 0$, we can assume that $x > 0$ when looking at the limit. If you are using the epsilon-delta definition, then the justification is that we can take $\delta < x_0$. So that $|x - x_0| < \delta \ \Rightarrow \ x > 0$. Does that make sense?

it's make sense , thank you

 

[El foolish Phenomeno]

Yes to both questions.

Not exactly. Since 1/2 and 2 are endpoints of intervals in which different formulas apply to get the two-sided limits at each of these points you'll need to determine the left- and right-side limits at each of these points, meaning that you will need to use the appropriate function formula for a left-side limit or a right-side limit.

BTW, your posted pictures are unreadable, at least by me. Your handwriting is very difficult to read, and the extremely poor lighting in the images make things worse. For reasons of legibility we discourage the use of images that show the work done.

next time i will try to write everything in latex. Thanks for the help

 

[e_jane]

There's no such thing as a "piecewise" function. It's just a regular function that is given by a different formula for different values of $x$. There are, therefore, no different rules.

A function which has different rules ("formulas") for different input values ("different values of $x$") is what is called a "piecewise function". In other words, a piecewise function is one which is defined by different rules for different pieces of the real number line. The term "piecewise function" is common in pre-calculus algebra (at least within the US, wherein lies most of my experience).

 

[PeroK]

A function which has different rules ("formulas") for different input values ("different values of $x$") is what is called a "piecewise function". In other words, a piecewise function is one which is defined by different rules for different pieces of the real number line. The term "piecewise function" is common in pre-calculus algebra (at least within the US, wherein lies most of my experience).

My point is that is a vacuous definition that has no mathematical content. A function is a function. It should be a "function defined piecewise". Hence the OP's confusion that they are different from regular functions, with different properties.

 

[e_jane]

A function is a function. It should be a "function defined piecewise". Hence the OP's confusion that they are different from regular functions, with different properties.

Good point. I wish textbooks and (more) instructors took greater care on this topic, precisely to avoid that confusion.