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As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.

    $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.

  2. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.

  3. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.

    $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.

  2. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.

  3. False in general, for example $f(x)=-g(x)=1$

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.
  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.
  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.

  2. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.

  3. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.

  2. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.

  3. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.

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user2139
user2139

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.
  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.
  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professorprofessors I'm not TA'ing for.

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.
  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.
  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professor I'm not TA'ing for.

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.
  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.
  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professors I'm not TA'ing for.

Source Link
user2139
user2139

What is a better way to explain these claims about limit are not true in general?

As a TA who led calculus* 1 and 2 discussion section and holds office hour** in the previous year, I heard the following (wrong) arguments several times.

  1. $\displaystyle \lim_{x\to \infty} \sqrt{x+1}-\sqrt{x}=0$ because $\infty-\infty=0$.
  1. $\displaystyle \lim_{x\to \infty} x^{1/x}=1$ because $\infty^0=1$.
  1. $\int_1^{\infty}f(x)dx$ and $\int_1^{\infty}g(x)dx$ both diverge so $\int_1^{\infty}f(x)+g(x)dx$ diverge.

I usually explain the arguments are not true in general by providing a (very trivial) counter example, for example,

  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=\infty$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)-g(x)=0$, for example, $f(x)=x+1$ and $g(x)=x$.
  1. $\displaystyle \lim_{x\to \infty} f(x)=\infty$ and $\displaystyle \lim_{x\to \infty} g(x)=0$ does not guarantee $\displaystyle \lim_{x\to \infty} f(x)^{g(x)}=1$, for example, $f(x)=2^x$ and $g(x)=1/x$.
  1. False in general, for example $f(x)=-g(x)=1$

After giving explanations like that I sometime heard "But in your examples you can cancel the expression/formula..." and I was not sure how to continue. I tried the following methods, non of them seem to work very well.

a. Provide a much more complicated counter example which requires a few minutes of calculation to get the answer. This often leads to further confusion.

b. Just say that is the wrong way to do it. It sounds like "I'm the teacher so believe me." and doesn't do too much.

c. Show them the correct way to do their problems. This is almost like b (Why is your way the right way and mine is the wrong way?).

I'm looking for a better way to deal with questions like these.

*$\epsilon-\delta$ definition is not introduced. ** Office hour is in tutoring center where I'm also responsible for students take the class from the professor I'm not TA'ing for.