# Thread: limit problems

1. ## limit problems

On Monday our teacher surprised us by telling us that we have a quiz on wednesday. Tonight she gave us homework, and I went through all of the questions, but there were ton of stuff I didn't see before. So I tried doing it on my own, but it got kinda crazy. So here they are. Sorry for not using latex, but I just don't got the time to learn the complex latex codes at the moment. I'll try to learn them after my midterm, which is next week.
1c) lim x-->0 (sin3X)/5X
e) lim x-->0 x cscx
g) lim x-->0 (1-cos2X)/X
i) lim x-->(pie/2) (((pie/2)-x)/(cosx)

2.Differentiate y with respect to X (I think we have to simplify which makes me )
2a)y= ((tan^4)(3x)
c)y=secx^2
d) y=(cot^2)2X/(1+x^2)
f) y=2sec(squareroot X)
h) y=cos^2(tanx)
i) y=1/(sin(x-sinx)

3. A triangle has adjacent sides of 4cm and 6cm. Prove that the traingle has a maximum area when the angle enclosed by these sides is 90degree.

4. A rain gutter is to be constructed wfrom a metal sheet of width 30 cm by bending up one third of the sheet on each side through an angle (theata). How should theata be chosen so the gutte will carry the maximum amount of water.
-----
On another sheet
A rectangular shet of iron 300 cm wide is to be bent to form a gutter whose cross section is the arc of a circle. What radius will give the gutter the maximum carrying capacity.

Thats all folks.

2. Hi

Originally Posted by Mr. Edward
1c) lim x--> (sin3X)/5X
e) lim x--> x cscx
g) lim x--> (1-cos2X)/X
i) lim x--> (((pie/2)-x)/(cosx)
$\displaystyle x\to ?$

For this four limits, think about the definition of the derivative : $\displaystyle f'(t)=\lim_{h\to0}\frac{f(t+h)-f(t)}{h}$

For example, for the first one (I guess $\displaystyle x\to 0$) :

$\displaystyle \lim_{x\to0}\frac{\sin(3x)}{5x}=\lim_{x\to0} \frac{1}{5}\cdot\frac{\sin(3(0+x))-\sin(3\cdot0)}{x}=\frac{1}{5}\cdot 3\cdot \cos(0)=\ldots$

because the derivative of $\displaystyle x\mapsto \sin(3x)$ is $\displaystyle x\mapsto3\cos(3x)$

3. Originally Posted by Mr. Edward
On Monday our teacher surprised us by telling us that we have a quiz on wednesday. Tonight she gave us homework, and I went through all of the questions, but there were ton of stuff I didn't see before. So I tried doing it on my own, but it got kinda crazy. So here they are. Sorry for not using latex, but I just don't got the time to learn the complex latex codes at the moment. I'll try to learn them after my midterm, which is next week.
1c) lim x--> (sin3X)/5X
e) lim x--> x cscx
g) lim x--> (1-cos2X)/X
i) lim x--> (((pie/2)-x)/(cosx)

2.Differentiate y with respect to X (I think we have to simplify which makes me )
2a)y= ((tan^4)(3x)
c)y=secx^2
d) y=(cot^2)2X/(1+x^2)
f) y=2sec(squareroot X)
h) y=cos^2(tanx)
i) y=1/(sin(x-sinx)

3. A triangle has adjacent sides of 4cm and 6cm. Prove that the traingle has a maximum area when the angle enclosed by these sides is 90degree.

4. A rain gutter is to be constructed wfrom a metal sheet of width 30 cm by bending up one third of the sheet on each side through an angle (theata). How should theata be chosen so the gutte will carry the maximum amount of water.
-----
On another sheet
A rectangular shet of iron 300 cm wide is to be bent to form a gutter whose cross section is the arc of a circle. What radius will give the gutter the maximum carrying capacity.

Thats all folks.
For the limit problems, it would help if you stated what x is approaching. I'm going to assume for the first limit problem that $\displaystyle x \to 0$. So what is the limit of $\displaystyle \frac{\sin{(3x)}}{5x}$ as x approaches 0?
Well, that expression is equal to:
$\displaystyle \frac{3}{5}\cdot \frac{\sin{(3x)}}{3x}$. And you should remember that the limit of $\displaystyle \frac{\sin{\theta}}{\theta}$ as theta approaches 0 is 1. Hence:
$\displaystyle \frac{3}{5}\cdot \frac{\sin{(3x)}}{3x} \to \frac{3}{5}$.

4. For the differentiation problems, use the chain rule:
$\displaystyle \frac{d}{dx} f(g(x)) = g'(x)\cdot f'(g(x))$.

I'll give you a hint on the first of these: $\displaystyle \frac{d}{dx} \tan{x} = \sec ^2{x}$.

5. I am sorry about that I fixed it now.

6. Originally Posted by Mr. Edward
1c) lim x-->0 (sin3X)/5X
e) lim x-->0 x cscx
g) lim x-->0 (1-cos2X)/X
i) lim x-->(pie/2) (((pie/2)-x)/(cosx)
You can use L'Hospital's rule (if this is unfamiliar to you, you should be able to find it by looking it up in the index of your book)

1c) $\displaystyle \lim_{x\to 0} \frac {sin(3x)}{5x} ~~~~=~~~~ \lim_{x\to 0} \frac {3cos(3x)}{5} ~~~~=~~~~ \frac 35 cos(0) ~~~~=~~~~ \frac 35$

e) $\displaystyle \lim_{x\to 0} x ~cscx ~~~~=~~~~ \lim_{x\to 0} \frac x{sin~x} ~~~~$ (use L'Hospital's here) $\displaystyle = \lim_{x\to 0} \frac 1{cos~x} ~~~~=~~~~ \frac x{cos 0} ~~~~=~~~~ 1$

i) $\displaystyle \lim_{x\to \pi /2} \frac {\pi /2 -x}{cos ~x}~~~~$ (use L'Hospitals here) $\displaystyle ~~ = \lim_{x\to \pi /2} \frac {-1}{-sin ~x} ~~~~=~~~~ \frac {1}{sin ~\pi /2}~~~~=~~~~1$

Originally Posted by Mr. Edward
2.Differentiate y with respect to X (I think we have to simplify which makes me )
2a)y= ((tan^4)(3x)
c)y=secx^2
d) y=(cot^2)2X/(1+x^2)
f) y=2sec(squareroot X)
h) y=cos^2(tanx)
i) y=1/(sin(x-sinx)[/tex]
These are applications of the chain rule

2a) $\displaystyle y = [tan(3x)]^4$

$\displaystyle y\prime = 4[tan(3x)]^3 * \left( \frac d{dx} tan(3x)\right)$

$\displaystyle y\prime = 4[tan(3x)]^3 * sec^2(3x) * \left(\frac d{dx}3x\right)$

$\displaystyle y\prime = 4[tan(3x)]^3 * sec^2(3x) * 3$

$\displaystyle y\prime = 12~tan^3(3x)~sec^2(3x)$

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For this one, I'm assuming you know that square root means to the power of 1/2, and that a power of -1/2 means 1 over the square root. I found especially when I was first doing derivatives, that keeping everything in fraction form (and using negative fractions instead of doing 1 over some expression) made things much simpler and prevented errors.

f) $\displaystyle y=2~sec(x^{1/2})$

$\displaystyle y\prime=2 ~sec(x^{1/2})~tan(x^{1/2}) * \left( \frac d{dx}x^{1/2}\right)$

$\displaystyle y\prime=2 ~sec(x^{1/2})~tan(x^{1/2}) * \frac 12 x^{-1/2} * \left(\frac d{dx} x\right)$

$\displaystyle y\prime=2 ~sec(x^{1/2})~tan(x^{1/2}) * \frac 12 x^{-1/2}$

$\displaystyle y\prime= \frac{sec(\sqrt{x})~tan(\sqrt{x})}{\sqrt{x}}$

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i) $\displaystyle y=\frac 1{sin(x-sin~x)}$

$\displaystyle y=[sin(x-sin~x)]^{-1}$

$\displaystyle y\prime=-[sin(x-sin~x)]^{-2} * \left( \frac d{dx} sin(x-sin~x) \right)$

$\displaystyle y\prime=-[sin(x-sin~x)]^{-2} * cos(x-sinx) \left( \frac d{dx} (x-sin~x) \right)$

$\displaystyle y\prime=-[sin(x-sin~x)]^{-2} * cos(x-sinx) ~(1-cos~x)$

$\displaystyle y\prime=-\frac{cos(x-sinx) ~(1-cos~x)}{sin^2(x-sin~x)}$

$\displaystyle y\prime=-\frac{cos(x-sinx) -cos~x~cos(x-sinx))}{sin^2(x-sin~x)}$

Was gonna do the others, but it just started storming :/