Use Taylor series to evaluate the following limits. lim_{xrightarrow0}frac{sqrt{1+2x}-1-x}{x^2}

Find the following:
${\displaystyle \underset{x\to -2}{lim}(\frac{1}{x^2}+\frac{1}{x^3+4})}$
Explain your answers using the continuity property of the addition, subtraction, multiplication, and division.

Evaluate the following limits. $\underset{(x,y)\to (e^2,4)}{lim}\ln \sqrt{xy}$

Using the trigonometric identity of $\sin 2\alpha =2\sin \alpha \cos \alpha$, I rewrote the expression to:
$\underset{n\to \mathrm{\infty }}{lim}(\frac{\sin (2\sqrt{1})}{n\sqrt{1}\cos \sqrt{1}}+\cdots +\frac{\sin (2\sqrt{n})}{n\sqrt{n}\cos \sqrt{n}})=\underset{n\to \mathrm{\infty }}{lim}(\frac{2\sin (\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\sin (\sqrt{n})}{n\sqrt{n}})$
I then tried using the squeeze theorem to get the limit, but I can't get an upper sequence that converges to zero, and the best I could get is a sequence that converges to two:
$\frac{2\sin (\sqrt{1})}{n\sqrt{1}}+\cdots +\frac{2\sin (\sqrt{n})}{n\sqrt{n}}\le \frac{2}{n\sqrt{1}}+\cdots +\frac{2}{n\sqrt{n}}\le \frac{2}{n}\cdot n⟶2$
What am I missing?

Evaluate the following limits. $\underset{x\to \mathrm{\infty }}{lim}(3\cdot 2^{1-x}+x^2\cdot 2^{1-x})$

I need help calculating this limit:
${\displaystyle \underset{x\to \pi }{lim}\frac{\cos 2x+2(\pi -x)\sin x-1}{{(\cos 2x+\cos x)}^2}}$
I tried to use L'Hospital's rule, but didn't get any result. Can you help me please?

Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
${\displaystyle f\left(x\right)={(x+2x^3)}^4,a=-1}$

How to evaluate
${\displaystyle \underset{x\to \mathrm{\infty }}{lim}\sqrt[n]{x^n+a_{n-1}{x}^{n-1}+\cdots +a_0}-x}$