Question

show that lim x->0 [(2+x)sin(2+x) - 2sin2]/x = 2cos2+sin 2

Answer 1

Let \[
f(x)= x \sin (x)\\
f'(x)=\sin (x)+x \cos (x)
\]
This is a hint. Compute f'(2) from the definition

Answer 2

\[
\frac{f(x+2)-f(2)}{x}=\frac{(x
+2) \sin (x+2)-2 \sin
(2)}{x}\\
\lim_{x->0} \frac{f(x+2)-f(2)}{x}=f'(2)=\sin (2)+2 \cos (2)
\]

Answer 3

@eliassaab we cannot use L'hospital

Answer 4

It is obvious that your limit is the derivative f'(2)

Answer 5

Alternatively, assuming we don't know how to find derivatives yet, you can use the angle sum formula for sine:
\[\sin(x+2)=\sin x\cos2+\cos x\sin 2\]
So,
\[\lim_{x\to0}\frac{(x+2)\sin(x+2)-2\sin 2}{x}=\lim_{x\to0}\frac{(x+2)(\sin x\cos2+\cos x\sin2)-2\sin 2}{x}\\
~~~~~~~~=\lim_{x\to0}\frac{x\sin x\cos2+2\sin x\cos2+x\cos x\sin2+2\cos x\sin2-2\sin2}{x}\\
~~~~~~~~=\cos2\left(\lim_{x\to0}\frac{\sin x(x+2)}{x}\right)+\sin2\left(\lim_{x\to0}\frac{x\cos x}{x}+2\lim_{x\to0}\frac{\cos x-1}{x}\right)
\]
which can be reduced if you know that
\[\lim_{x\to0}\frac{\sin ax}{ax}=1~~~\text{and}~~~\lim_{x\to0}\frac{1-\cos ax}{ax}=0~~~\text{for }a\not=0\]

Answer 6

@SithsAndGiggles , your method is correct. It is longer and you have to find the derivative anyway.

Answer 7

@SithsAndGiggles how do we show the last step
ie: 2cos2+sin 2

Answer 8

ok i got it @SithsAndGiggles

Answer 9

Thanks for your help !!

Answer 10

\[\cos2\left(\lim_{x\to0}\frac{\sin x(x+2)}{x}\right)+\sin2\left(\lim_{x\to0}\frac{x\cos x}{x}+2\lim_{x\to0}\frac{\cos x-1}{x}\right)\]
One of the properties of limits you should be aware of is that the limit of a product is the same as the product of the limits; mathematically, you have
\[\lim_{x\to c}f(x)g(x)=\left(\lim_{x\to c}f(x)\right)\left(\lim_{x\to c}g(x)\right)\]
You can apply this property directly to the first limit:
\[\cos2\left(\lim_{x\to0}\frac{\sin x(x+2)}{x}\right)=\cos2\left(\color{red}{\lim_{x\to0}\frac{\sin x}{x}}\right)\left(\lim_{x\to0}(x+2)\right)\]
In the second limit, the \(\dfrac{x}{x}\) just becomes 1, so you're left with \(\cos x\). The third limit is known.

Answer 11

You're welcome!

Answer 12

Here is a point that many people do not know. To prove that the derivative of sin(x) is cos(x), we use the geometric fact that sin(x)/x approaches 1 when x approaches zero.
You really cannot use L'hospital rule to show that that sin(x)/x approaches 1 when x approaches zero, but people do. C'est la vie.