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

Can someone please help me with calculus , i am having so much trouble. Thank you-example-1

1 Answer

6 votes

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.

User Matousc
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