Limits and Continuity

[LadyJonesVII]

Hello, I am currently taking a Calc 1 course and I am having some difficulty on this section about Continuity of Trigonometric Functions. Sometimes I am unsure where to begin; learning trigonometric identities is important but I do not know how to apply them in certain cases. Or know the first steps I should take when trying to find limits.

Ex. Find the limits.

lim x[sup:1ogxn9f3]2[/sup:1ogxn9f3] - 2 sin x/x
x----> 0

lim tan 3x[sup:1ogxn9f3]2[/sup:1ogxn9f3] + sin[sup:1ogxn9f3]2[/sup:1ogxn9f3] 5x/ x[sup:1ogxn9f3]2[/sup:1ogxn9f3]
x----> 0

Any help is greatly appreciated, thank-you.

 

[tkhunny]

First, let's be more clear with notation.

Do you mean \(\displaystyle x^{2} - \frac{2\cdot \sin(x)}{x}\), as you have written, or do you mean \(\displaystyle \frac{x^{2}-2\sin(x)}{x}\)?

Likewise for the second.

 

[renegade05]

Note: Sorry, I guess I hit "Edit" rather than "Quote". This was not my intent. I did object to one premise, but your advice on sin(x)/x was right on point.

Do those not produce the same answer?

Only by coincidence. Never rely on coincdence.

For the OP, simply try adding parentheses. These are not the same

x - a / 2

(x - a) / 2

You don't have to code all fancy, just write clearly.

 

[LadyJonesVII]

First, let's be more clear with notation.

Do you mean \(\displaystyle x^{2} - \frac{2\cdot \sin(x)}{x}\), as you have written, or do you mean \(\displaystyle \frac{x^{2}-2\sin(x)}{x}\)?

Likewise for the second.

I mean the second one you wrote, I did not mean to cause any misunderstandings. If I could get it to display as you have done then it would have been more clear, but I have no idea how to do that.

@renegade05

Right I remember that, so does this mean that basically it is x[sup:1qi71y77]2[/sup:1qi71y77]- 3 times 1? And that zero is substituted for the x[sup:1qi71y77]2[/sup:1qi71y77], thus it becomes 0-3 which leaves me with -3 for my first answer? So then what do I apply, or have to do, for the second question?

 

[renegade05]

Note:Sorry, i guess I hit "Edit" rather than "Quote". This was not my intent. I did object to one premise, but your advice on sin(x)/x was right on point.

Ya, that is OK. At first glance, i was like .. "jeez i know I am new here but to delete me post... " But i got you now.

Right I remember that, so does this mean that basically it is x[sup:3b67sj91]2[/sup:3b67sj91]- 3 times 1? And that zero is substituted for the x[sup:3b67sj91]2[/sup:3b67sj91], thus it becomes 0-3 which leaves me with -3 for my first answer? So then what do I apply, or have to do, for the second question?

I think you made a typo there. It is not -3.. but...

For the second try to break it up into two limits

\(\displaystyle \lim_{x \to 0}tan3x^2 + \lim_{x \to 0}\frac{\sin^25x}{x^2}\)

I am assuming this is your problem?? I cant tell for the way you have written it.

 

[LadyJonesVII]

Really? Ha, for the answer it says it is -3 for the first one. And alright I'll just use parenthesis next time for clarification. So anyway the second problem was \(\displaystyle \lim_{x \to 0}\frac{tan3x^2 +\sin^25x}{x^2}\)
And yes, I see that I would have to break the limits up and figured out how to solve the problem.

My Steps for the second problem:

\(\displaystyle \lim_{x \to 0}\frac{tan3x^2}{x^2} + \lim_{x \to 0}\frac{\sin^25x}{x^2}\)

\(\displaystyle 3 + \lim_{x \to 0}(\frac{\sin5x}{x^2})^2\)

\(\displaystyle =3+25\)

\(\displaystyle =28\)

 

[mmm4444bot]

Really? Yes.

the answer [key] says [the limit] is -3 for the first [exercise].

The answer key is wrong.

:idea: If you have graphing technology, it's easy to confirm their error by graphing [x^2 - 2 sin(x)]/x and looking at the y-intercept.



alright I'll just use [parentheses] next time for clarification.

That's a great plan because it will put you ahead of the "game".



for the second problem:

\(\displaystyle 3 + \lim_{x \to 0}(\frac{\sin5x}{x^2})^2\)

\(\displaystyle = 28\)

This second limit is 28, so I'm thinking that your x^2 in the denominator above is a typographical error.