Question

Find the limit if they exist:
f(x) = \[\frac{ asin(x) - xsin(a) }{ x - a } x < a\]
f(x) = \[\frac{ \sin(x) - \sin(a) }{ x - a } x > a\]

Answer 1

1. Doesn't exist
2. Doesn't exist

Answer 2

Steps ?

Answer 3

Are these two separate functions, or is it
\[f(x)=\begin{cases}\dfrac{a\sin x-x\sin a}{x-a}&\text{for }x<a\\\\
\dfrac{\sin x-\sin a}{x-a}&\text{for }x>a\end{cases}\]

Answer 4

It's like you typed it.

Answer 5

Also is \(x\to a\) from a particular direction? Or are you asked to find the one-sided limits only?

Answer 6

Two separate functions I believe.

Answer 7

Not separated bro.

Answer 8

Oh its not :o

Answer 9

You'ill have to find the limit from both sides so as to compare the values if equal it exists

Answer 10

Can you solve it ?

Answer 11

Sorry I have no idea, I tried.

Answer 12

The right-sided limit is fairly simple. Recall the definition of the derivative:
\[f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\]
This means that
\[\large\lim_{x\to a^+}f(x)=\lim_{x\to a^+}\frac{\sin x-\sin a}{x-a}=g'(a)\]
where \(g(x)=\sin x\). Hence the right-sided limit is \(\cos a\), since \(\dfrac{d}{dx}\sin x=\cos x\).

Answer 13

For the left-sided limit, you have
\[\begin{align*}\frac{a\sin x-x\sin a}{x-a}&=\frac{a\sin x-(x-a+a)\sin a}{x-a}\\\\
&=\frac{a\sin x-a\sin a-(x-a)\sin a}{x-a}\\\\
&=a\left(\frac{\sin x-\sin a}{x-a}\right)-\sin a
\end{align*}\]
As before, you'll find that the fraction approaches \(\cos a\), and so the right-sided limit is \(a\cos a-\sin a\).

Answer 14

Cool dude , i really like when you + a and - it again !

But I have a small difficulty to understand from "This mean that to" in the first comment the second one is understandable

Answer 15

I think what's confusing you is that I'm using \(f\) to mean two different things.

The first \(f\), in the definition of the derivative, is a general differentiable function. The second one is the piecewise function given above.

Let me write it another way.
\[F'(c)=\lim_{h\to0}\frac{F(c+h)-F(c)}{h}=\lim_{x\to c}\frac{F(x)-F(c)}{x-c}\]
Denote the piecewise function as \(f\), then
\[\lim_{x\to a^+}f(x)=\lim_{x\to a^+}\frac{\sin x-\sin a}{x-a}=\lim_{x\to a^+}\frac{F(x)-F(a)}{x-a}=F'(a)\]
Then, since \(F(x)=\sin x\) and \(F'(x)=\cos x\), you get
\[\lim_{x\to a^+}f(x)=\lim_{x\to a^+}\frac{\sin x-\sin a}{x-a}=\lim_{x\to a^+}\frac{F(x)-F(a)}{x-a}=\cos a\]

Answer 16

Thanks ! fantastic, really a masterpiece.
Could you help me with another one , if you don't mind ?