# Thread: help with 2 questions

1. ## help with 2 questions

i have problems with those questions?

2. Originally Posted by sbsite
i have problems with those questions?
To find the inverse function we switch x and y and then solve for y:

$\displaystyle x = \frac{e^Y + e^{-Y}}{2}$

Let $\displaystyle n = e^Y$. Then $\displaystyle \frac{1}{n} = e^{-Y}$.

So
$\displaystyle x = \frac{n + \frac{1}{n}}{2}$

$\displaystyle 2x = n + \frac{1}{n}$

$\displaystyle 2xn = n^2 + 1$

$\displaystyle n^2 - 2xn + 1 = 0$

$\displaystyle n = \frac{2x \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Now, $\displaystyle Y = ln(n)$

$\displaystyle Y = ln \left ( x \pm \sqrt{x^2 - 1} \right )$

Since the argument of ln can never be negative, we choose the "+" sign.
$\displaystyle Y = ln \left ( x + \sqrt{x^2 - 1} \right )$

-Dan

3. Originally Posted by sbsite
i have problems with those questions?
$\displaystyle \lim_{h \to 0}\frac{(2+h)^4 - 16}{h}$

= $\displaystyle \lim_{h \to 0}\frac{16 + 32h + 24h^2 + 8h^3 + h^4 - 16}{h}$

= $\displaystyle \lim_{h \to 0}\frac{32h + 24h^2 + 8h^3 + h^4}{h}$

= $\displaystyle \lim_{h \to 0}\frac{h(32 + 24h + 8h^2 + h^3)}{h}$

= $\displaystyle \lim_{h \to 0}(32 + 24h + 8h^2 + h^3)$

= 32

-Dan

4. Originally Posted by sbsite
i have problems with those questions?
$\displaystyle \lim_{x \to 1} \frac{\sqrt{x}}{x+1}$

$\displaystyle \lim_{x \to 2} \frac{x^2 - 1}{x - 1}$

Neither of these is difficult. In both cases the numerator and denominator functions are non-zero and continuous at the limit points. So just plug in the x values.

-Dan

5. Hello, sbsite!

I think I have the last part of #12 . . .

$\displaystyle (1)\;\lim_{x\to0^+}\frac{1 + 10^{-\frac{1}{x}}}{2 - 10^{-\frac{1}{x}}} \qquad (2)\;\lim_{x\to0^-}\frac{1 + 10^{-\frac{1}{x}}}{2 - 10^{-\frac{1}{x}}} \qquad (3)\;\lim_{x\to0}\frac{1 + 10^{-\frac{1}{x}}}{2 - 10^{-\frac{1}{x}}}$

Let $\displaystyle y \,= \,\frac{1}{x}$ . . When $\displaystyle \begin{array}{cc}x\to0^+,\;y\to+\infty \\ x\to0^-,\;y\to-\infty\end{array}$

(1) We have: .$\displaystyle \lim_{y\to\infty}\frac{1 + 10^{-y}}{2 - 10^{-y}} \;=\;\lim_{y\to\infty}\frac{1 + \frac{1}{10^y}}{2 - \frac{1}{10^i}} \;=\;\frac{1 + 0}{2 - 0} \;=\;\boxed{\frac{1}{2}}$

(2) We have: .$\displaystyle \lim_{y\to\text{-}\infty}\frac{1 + 10^{-y}}{2 - 10^{-y}}$

. . Multiply top and bottom by $\displaystyle 10^y\!:\;\;\lim_{y\to\text{-}\infty}\frac{10^y + 1}{2\!\cdot\!10^y - 1} \;=\;\frac{0 + 1}{2\!\cdot\!0 - 1} \;=\;\boxed{-1}$

(3) The left- and right-hand limits are unequal.
. . Therefore: .$\displaystyle \lim_{x\to0}\frac{1 + 10^{-\frac{1}{x}}}{2 - 10^{-\frac{1}{x}}}$ does not exist.

6. ## help??

[B]I still have problems with the two questions that indicated in red??
I need to solve this assignment until Sunday?
help??[/
B]
thanks

7. Originally Posted by sbsite
[b]I still have problems with the two questions that indicated in red??
I need to solve this assignment until Sunday?
help??[/B]
thanks
$\displaystyle \lim_{x \to 0} x^2 cos \left ( \frac{1}{x} \right )$

I think we can do this in a straightforward manner, but let's borrow Soroban's trick. I think it makes things a touch clearer:

Let $\displaystyle y = \frac{1}{x}$. Then
$\displaystyle \lim_{x \to 0} x^2 cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{1}{y^2}cos(y)$

Now, even though $\displaystyle \lim_{y \to \infty} cos(y)$ doesn't exist, we know that the cosine function only varies between -1 and 1. So we can pretend that the limit does exist and is some finite number between -1 and 1. BUT $\displaystyle \lim_{y \to \infty} \frac{a}{y^2} = 0$ for any finite a. Thus:
$\displaystyle \lim_{x \to 0} x^2 cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{1}{y^2}cos(y) = 0$

-Dan

8. Originally Posted by sbsite
[b]I still have problems with the two questions that indicated in red??
I need to solve this assignment until Sunday?
help??[/B]
thanks
Explaining this comes in two steps.

1. Translation. Given the graph of y = f(x), we know that the graph of y' = f(x - h) is a translation of the original graph by h units to the right. Similarly y' = f(x + h) is a translation of the original graph by h units to the left.

2. Dilation (aka "stretching" and "shrinking"). Given the graph y = f(x), we know that the graph of y' = f(ax) for a > 1 "shrinks" the x-axis by a factor of a. (The graph becomes thinner.) For y'' = f(ax) and 0 < a < 1 the x-axis is "stretched" by a factor of 1/a. (The graph becomes wider.)

So.
For y = cos(3x), take the graph of y = cos(x) and shrink the axis by a factor of 3. That is, instead of the function having a period of $\displaystyle 2\pi$ it will now have a period of $\displaystyle \frac{2\pi}{3}$, so the interval $\displaystyle [0, 2\pi)$ now represents 3 oscillations.

For the other problem, you could simply take the graph of y = sin(x), shrink the x-axis by a factor of 2, then translate the graph $\displaystyle \frac{\pi}{3}$ units to the left. However, I personally prefer to do my dilations last, so I'd do the following:

$\displaystyle y = sin \left ( 2x + \frac{\pi}{3} \right )$

$\displaystyle y = sin \left ( 2x + 2 \frac{\pi}{6} \right )$

$\displaystyle y = sin \left ( 2 \left [ x + \frac{\pi}{6} \right ] \right )$

So we first want to take the graph of y = sin(x) and translate it $\displaystyle \frac{\pi}{6}$ units to the left, then shrink the x-axis by a factor of 2.

-Dan

9. Originally Posted by topsquark
$\displaystyle \lim_{x \to 0} x^2 cos \left ( \frac{1}{x} \right )$
You should use the squeeze theorem,
$\displaystyle |\cos (1/x)|\leq 1$
Thus,
$\displaystyle -x^2\leq x^2\cos (1/x) \leq x^2$
For some open interval containing x=0 except possibly at x=0.

10. Originally Posted by ThePerfectHacker
You should use the squeeze theorem,
$\displaystyle |\cos (1/x)|\leq 1$
Thus,
$\displaystyle -x^2\leq x^2\cos (1/x) \leq x^2$
For some open interval containing x=0 except possibly at x=0.
I agree, much better. I always forget about that one.

-Dan