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Solution:

Part a

We have $\lim_{x\to 0}(e^{2x}-\cos x)=1-1=0$ and $\lim_{x\to 0} x=0$. Hence we can apply L’Hospital’s Rule and obtain that
\begin{align*}
\lim_{x\to 0}\frac{e^{2x}-\cos x}{x}
=&\ \lim_{x\to 0}\frac{(e^{2x}-\cos x)’}{(x)’}\\
=&\ \lim_{x\to 0}\frac{2e^{2x}+\sin x}{1}\\
=&\ \lim_{x\to 0}(2e^{2x}+\sin x)=2+0=2.
\end{align*}

Part b

We have $\lim_{x\to 0}(1-\cos x)=1-1=0$ and $\lim_{x\to 0} x^2=0$. Hence we can apply L’Hospital’s Rule and obtain that
\begin{align*}
\lim_{x\to 0}\frac{1-\cos x}{x^2}
=&\ \lim_{x\to 0}\frac{(1-\cos x)’}{(x^2)’}\\
=&\ \lim_{x\to 0}\frac{\sin x}{2x}\\
\text{ Apply L’Hospital’s Rule again }\ =&\ \lim_{x\to 0}\frac{\cos x}{2}=\frac{1}{2}.
\end{align*} Here we can apply L’Hospital’s Rule again because $\lim_{x\to 0}\sin x=0$ and $\lim_{x\to 0}2x=0$.

Part c

Whenever the denominator tends to infinity, we can apply L’Hospital’s Rule. Therefore, we can repeatedly apply L’Hospital’s Rule as follows
\begin{align*}
\lim_{x\to \infty}\frac{x^3}{e^{2x}}
=&\ \lim_{x\to \infty}\frac{3x^2}{2e^{2x}}\\
\text{ Apply L’Hospital’s Rule again }\ =&\ \lim_{x\to \infty}\frac{6x}{4e^{2x}}\\
\text{ Apply L’Hospital’s Rule again }\ =&\ \lim_{x\to \infty}\frac{6}{8e^{2x}}=0.
\end{align*}

Part d

We have $\lim_{x\to 0}(\sqrt{1+x}-\sqrt{1-x})=1-1=0$ and $\lim_{x\to 0} x=0$. Hence we can apply L’Hospital’s Rule and obtain that
\begin{align*}
\lim_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}
=&\ \lim_{x\to 0}\frac{(\sqrt{1+x}-\sqrt{1-x})’}{(x)’}\\
=&\ \lim_{x\to 0}\frac{ \frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}} }{1}\\
=&\ \lim_{x\to 0}\Big(\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}\Big)\\
=&\ \frac{1}{2}+\frac{1}{2}=1.
\end{align*}