Question

Can someone please help me with calculus , i am having so much trouble. Thank you! 12points

Answer 1

1) If the limit
`L` is

`L = \displaystyle \lim_(\Delta x\to0) (\sin\left(\frac\pi3 + \Delta x\right) - \sin\left(\frac\pi3\right))/(\Delta x)`

then using the hint as well as
`\sin\left(\frac\pi3\right)=\frac{\sqrt3}2` and
`\cos\left(\frac\pi3\right)=\frac12` we have

`\displaystyle L = \lim_(\Delta x\to0) (\sin\left(\frac\pi3\right)\cos(\Delta x) + \cos\left(\frac\pi3\right)\sin(\Delta x) - \sin\left(\frac\pi3\right))/(\Delta x)`

`\displaystyle L = \sin\left(\frac\pi3\right) \lim_(\Delta x\to0) (\cos(\Delta x) - 1)/(\Delta x) + \cos\left(\frac\pi3\right) \lim_(\Delta x) (\sin(\Delta x))/(\Delta x)`

`L = \frac{\sqrt3}2 * 0 + \frac12*1 = \boxed{\frac12}`

which follows from the well-known limits,

`\displaystyle \lim_(x\to0) \frac{1-\cos(x)}x = 0 \text{ and } \lim_(x\to0) \frac{\sin(x)}x=1`

Alternatively, if you already know about derivatives, we can identify the limit as the derivative of
`\sin(x)` at
`x=\frac\pi3`, which is
`\cos\left(\frac\pi3\right)=\frac12`.

2) It looks like you may be using double square brackets deliberately to denote the greatest integer or floor function which rounds the input down to the nearest integer. That is,
`[\![x]\!]` is the greatest integer that is less than or equal to
`x`. The existence of
`L` depends on the equality of the one-sided limits.

Suppose
`3\le x<4`. Then
`2[tex]\displaystyle \lim_(x\to4^-) [\![x - 1]\!] = 2`

Now suppose
`4\le x<5`, so that
`3\le x-1 < 4 \implies [\![x-1]\!]=3` and

`\displaystyle \lim_(x\to4^+) [\![x-1]\!] = 3`

The one-sided limits don't match so the limit doesn't exist.