I would appreciate if someone could verify to me my answers.
$$\lim_{x\rightarrow0}\frac{x-\sin(x)}{\tan(x)-x}$$I used L'Hopital's rule twice and got answer $1/2$. $$\lim_{x\rightarrow0^+}\frac{1}{\sin(4x)}-\frac{1}{4x}$$ also I used L'Hopital's rule twice and got $0$. $$\lim_{x\rightarrow2^-}(x^2-4)\ln(2-x)$$I used L'Hopital's rule once and got $0$.
Thanks.
Your answer is indeed correct. We can use the power series for sinx and tanx to verify the answer. We have $$\sin(x)=x-x^3/3!+x^5/5!...$$ and $$\tan(x)=x+x^3/3+2x^5/15...$$ Substituting the above two series in your limit, we get $$\lim_{x\rightarrow0}\frac{x-\sin(x)}{\tan(x)-x}$$ $$\lim_{x\rightarrow0}\frac{x-(x-x^3/3!+x^5/5!...)}{(x+x^3/3+2x^5/15...)-x}$$ Cancelling the x we get $$\lim_{x\rightarrow0}\frac{x^3/3!-x^5/5!...}{x^3/3+2x^5/15...}$$ Dividing numerator and denominator by $x^3$ $$\lim_{x\rightarrow0}\frac{1/3!-x^2/5!+...}{1/3+2x^2/15+...}$$ Taking limits the x vanishes and you are left with $$\frac{1/3!}{1/3}$$ Simplify to get $1/2$
Since the answer to the first problem has been given by GTX OC, let us focus on the second problem.
$$(x^2-4) \log(2-x) = (x-2)(x+2)\log(2-x)=-(x+2)(2-x)\log(2-x)$$
I suppose you know that $x \log(x)$ goes to 0 when $x$ goes to zero. Then, ... Are you able to continue with this ?
