Thread: Questions on Beginner Calculus Problems..

1. Questions on Beginner Calculus Problems..

I have three questions from my calculus homework that have confused me..

First.. I've been trying to evaluate this limit but it keeps messing me up.

Second, this confused me, also..
Let f(x) = . Is the function f(x) continuous at x = 0? Can the function be extended to that it is continuous?

Lastly..this one is just worded oddly..

Show that the line y = mx + b is its own tangent at any point
().

If any one could help me on any\all of this, I'd be very grateful ^_^ By the way, we haven't learned real derivatives yet so we have to do everything the long way in our work..

2. Originally Posted by starryemmy
I have three questions from my calculus homework that have confused me..

First.. I've been trying to evaluate this limit but it keeps messing me up.
Recall, $e^x$ is a continuous function.

So, $\lim_{x \to 0}e^{x \sin (1/x)} = e^{\lim_{x \to 0}x \sin (1/x)}$

so the real problem is to find $\lim_{x \to 0}x \sin \left( \frac 1x \right)$

Hint: You can use the squeeze theorem here.

Can you take it from here?

Second, this confused me, also..
Let f(x) = . Is the function f(x) continuous at x = 0? Can the function be extended to that it is continuous?
Recall the special limit: $\lim_{x \to 0} \frac {\sin x}x = 1$

also recall, $|x| = \left \{ \begin {array}{cc} x & \mbox {if } x \ge 0 \\ {-x} & \mbox {if } x < 0 \end {array} \right.$

Now, we note that: $\frac {\sin x}{|2x|} = \frac {\sin x}{2|x|} = \frac 12 \frac {\sin x}{|x|}$

since we have absolute values, we know that the $x$ can be negative or positive, depending on which direction we approach 0 from. Thus, when we see absolute values and we want the limit as we are approaching zero we must consider the one-sided limits.

So, $\lim_{x \to 0} \frac {\sin x}{|2x|} = \frac 12 \lim_{x \to 0} \frac {\sin x}{|x|}$ exists if and only if $\lim_{x \to 0^{+}} \frac {\sin x}{|x|}$ and $\lim_{x \to 0^{-}} \frac {\sin x}{|x|}$ exists and are equal.

now find the limit.

why do we need to find the limit? they asked us about continuity.

i'm glad you asked. recall what it means for a function to be continuous at a certain point. (i'll give you the watered-down version).

A function $f(x)$ is continuous at a point $a$ if:

$\lim_{x \to a}f(x) = f(a)$

Now continue...

Lastly..this one is just worded oddly..

Show that the line y = mx + b is its own tangent at any point
().

If any one could help me on any\all of this, I'd be very grateful ^_^ By the way, we haven't learned real derivatives yet so we have to do everything the long way in our work..
I need to know how far you have reached with derivatives. do you have to use the limit definition, or can you do it the short way?

3. Ah, for the last one we have to do it the long way with the definition

4. Originally Posted by starryemmy
Ah, for the last one we have to do it the long way with the definition
ok. what about the other questions? did my hints help?

so for the last question. remember, if the slope of the tangent line (which is given by the derivative) at a certain point (let's call it $x_0$) of a function is $m$. then the equation of the tangent line is given by the point-slope form:

$y - y_0 = m(x - x_0)$, where $(x_0,y_0)$ is the point we are concerned with.

so your first task is to find $m$ so you can plug it into that formula. you can find $m$ by the definition of the derivative (using these variables):

at $(x_0,y_0)$, $m = f'(x_0) = \lim_{h \to 0} \frac {f(x_0 + h) - f(x_0)}h$

(recall here, that $f(x_0) = y = mx_0 + b$)

when you find $m$, plug it into the point-slope formula to find the equation of the tangent line.

5. Originally Posted by Jhevon
so the real problem is to find $\lim_{x \to 0}x \sin \left( \frac 1x \right)$

Hint: You can use the squeeze theorem here.
Well, we can avoid the squeeze theorem by setting $u=\frac1x\implies u\to\infty$, the limit becomes to

$\lim_{x\to0}x\sin\frac1x=\lim_{u\to\infty}\frac{\s in u}u=0$

Mmm... though in the proof of $\lim_{x\to\infty}\frac{\sin x}x=0$ I'll use the squeeze theorem anyway

--

We know that $-1\le\sin x\le1,\,\forall x\in\mathbb R$

Let's take $x>0$. Now multiplyin' the previous inequality by $\frac1x$ yields

$-\frac1x\le\frac{\sin x}x\le\frac1x$. Now takin' the limit when $x\to\infty$, we can conclude that $\lim_{x\to\infty}\frac{\sin x}x=0$

We can take the case when $x<0$, yielding the same result but with a similar procedure.

6. Originally Posted by Krizalid
...

Mmm... though in the proof of $\lim_{x\to\infty}\frac{\sin x}x=0$ I'll use the squeeze theorem anyway
...
yeah, that's why i said to just use it to begin with. or maybe the proof of the squeeze theorem is not necessary. if the poster was taught the squeeze theorem in class, especially if it was proven, i think he can just use it

7. Okies, for my first problem I did get as far as seeing that I had to find the limit as x approaches 0 of xsin(1/x). Our professor skimmed through the steps for this though, so I was a bit lost. If you could explain, that'd be great.

For my second problem..

So the limit of f(x) as x approches 0 from the right and left equals 1, this doesn't make it continuous because f(1) doesn't equal 0? or is it the other way around..hm

The last problem I got m = m somehow. Is that correct?

8. Originally Posted by starryemmy
Okies, for my first problem I did get as far as seeing that I had to find the limit as x approaches 0 of xsin(1/x). Our professor skimmed through the steps for this though, so I was a bit lost. If you could explain, that'd be great.

For my second problem..

So the limit of f(x) as x approches 0 from the right and left equals 1, this doesn't make it continuous because f(1) doesn't equal 0? or is it the other way around..hm

The last problem I got m = m somehow. Is that correct?
that's what you should get, but m = m looks awkward and redundant. i should have said that f'(x_0) = ... so you would have ended up with f'(x_0) = m

now continue with that

9. y - y_0 = m (x - x_0)

is what I ended up with..but I can't do much else >.<

10. Originally Posted by starryemmy
y - y_0 = m (x - x_0)

is what I ended up with..but I can't do much else >.<
solving for y you get y = mx + (y_0 - mx_0), this is the tangent line, which is of the form y = mx + b

what can we say from this?