limits: lim[h->0] [sin(3x+h)-sin(3x)] / h

[riocean17]

lim[h->0] [sin(3x+h)-sin(3x)] / h

I'm having a hard time solving this. I can't even remember where to start. Can anyone get me started please?

Thank you!

 

[galactus]

Let's see. We could use L'Hopital's Rule, but that'd be too easy.

Let's use the limits: \(\displaystyle \L\\\lim_{h\to\infty}\frac{sin(h)}{h}=1 \;\ and \;\ \lim_{h\to\0}\frac{1-cos(h)}{h}=0\)


\(\displaystyle \L\\\lim_{h\to\0}\frac{sin(3x+h)-sin(3x)}{h}\)

\(\displaystyle \L\\\lim_{h\to\0}\frac{\overbrace{sin(3x)cos(h)+cos(3x)sin(h)}^{\text{addition rule for sin(u+v)}}-sin(3x)}{h}\)

Factor:

\(\displaystyle \L\\\lim_{h\to\0}\left[cos(3x)\left(\frac{sin(h)}{h}\right)-sin(3x)\left(\frac{1-cos(h)}{h}\right)\right]\)

Since cos(x) and sin(x) do not involve h, they remain as constants as h approaches 0.

So, we have:

\(\displaystyle \L\\cos(3x)\cdot\lim_{h\to\0}\left(\frac{sin(h)}{h}\right)-sin(3x)\cdot\lim_{h\to\0}\left(\frac{1-cos(h)}{h}\right)\)

\(\displaystyle \L\\cos(3x)(1)-sin(3x)(0)=cos(3x)\)


For L'Hopital, the derivative of the numerator is: \(\displaystyle cos(3x+h)\)

Therefore, you have:

\(\displaystyle \L\\\lim_{h\to\0}\frac{cos(3x+h)}{1}=cos(3x)\)

Much easier, but a bugaboo of mine is not to use L'Hopital unless I have to.

 

[ChaoticLlama]

Isn't the derivative of sin(3x), 3cos(3x)?

 

[galactus]

Yes, you're certainly correct. The chain rule would dictate.

\(\displaystyle \L\\\lim_{h\to\0}\frac{3(x+h)-3x}{h}=3\)

 

[pka]

Isn't the derivative of sin(3x), 3cos(3x)?

This problem is not about the derivative of sin(3x). I thought that at first.
But the limit for that derivative is \(\displaystyle \frac{{\sin \left[ {3\left( {x + h} \right)} \right] - \sin \left[ 3x \right]}}{h}\) and not \(\displaystyle \frac{{\sin \left[ {3x + h} \right] - \sin \left[ 3x \right]}}{h}\).

 

[galactus]

Oh, good catch, pka.