Find the following limits using Maclaurin series and check your results by computer. Hint: First combine the fractions. Then find the first term of the denominator series and the first term of the numerator series.

$$\begin{gathered} \left(a\right)\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{}\mathbf{lim}}\left(\frac{1}{x}-\frac{1}{e^x-1}\right) \\ \left(b\right)\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left(\frac{1}{x^2}-\frac{\cos x}{{\sin }^{2}x}\right) \\ \left(c\right)\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left({\csc }^{2}x-\frac{1}{x^2}\right) \\ \left(d\right)\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\left(\frac{\mathrm{In}\left(1+x\right)}{x^2}-\frac{1}{x}\right) \end{gathered}$$

The Maclaurin series, i.e.,$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(0\right)+{\mathrm{f}}^{\text{'}}\left(0\right)\mathrm{x}+\frac{{\mathrm{f}}^{"}\left(0\right)}{2!}{\mathrm{x}}^2+\frac{{\mathrm{f}}^{\left(3\right)}\left(0\right)}{3!}{\mathrm{x}}^3+...+\frac{{\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)}{\mathrm{n}!}{\mathrm{x}}^{\mathrm{n}}+....$

As the number of terms approaches infinity, the series' limit is the value that the series' terms are approaching.

Use Maclaurin series expansion of $e^x$,

$$\begin{gathered} e^x=1+x\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \\ {e}^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \\ {e}^x-1-x=\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \\ x(e^x-1)=x^2+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+... \end{gathered}$$

Further, solve the limit $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{x}-\frac{1}{e^x-1}\right)$as:

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\left(\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}}{\mathrm{x}\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{{\displaystyle \frac{{\mathrm{x}}^2}{2!}}+{\displaystyle \frac{{\mathrm{x}}^3}{3!}}+{\displaystyle \frac{{\mathrm{x}}^4}{4!}}+..}{{\mathrm{x}}^2+{\displaystyle \frac{{\mathrm{x}}^2}{2!}}+{\displaystyle \frac{{\mathrm{x}}^4}{3!}}+{\displaystyle \frac{{\mathrm{x}}^5}{4!}}+..}\right) \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{{\mathrm{e}}^{\mathrm{x}}-1-\mathrm{x}}{\mathrm{x}\left({\mathrm{e}}^{\mathrm{x}}-1\right)}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{\mathrm{x}}{3!}}+{\displaystyle \frac{{\mathrm{x}}^2}{4!}}+...}{1+{\displaystyle \frac{\mathrm{x}}{2}}+{\displaystyle \frac{{\mathrm{x}}^2}{3!}}+{\displaystyle \frac{{\mathrm{x}}^3}{4!}}+...}\right) \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{e}}^{\mathrm{x}}-1}\right)=\frac{1}{2} \end{gathered}$$

Use the MATLAB tool to verify the limit.

Hence, the required value of limits is mentioned as,$\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{e}}^{\mathrm{x}}-1}\right)=\frac{1}{2}$.

Use Maclaurin series expansion of $\sin \left(x\right)$,

$$\begin{gathered} \sin \left(x\right)=x-\frac{x^3}{6}+\frac{x^5}{120}+... \\ {\sin }^2\left(x\right)=\left[x-\frac{x^3}{6}+\frac{x^5}{120}+...\right].\left[x-\frac{x^3}{6}+\frac{x^5}{120}+...\right] \\ {\sin }^2\left(x\right)=x^2-\frac{x^4}{3}+\frac{2x^6}{45}+..... \end{gathered}$$

$$\begin{gathered} \cos \left(x\right)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+.... \\ {x}^2\cos \left(x\right)=x^2-\frac{x^4}{2}+\frac{x^6}{24}+.... \\ {\sin }^2\left(x\right)-x^2\cos \left(x\right)=\left[x^2-\frac{x^4}{3}+\frac{2x^6}{45}+....\right]-\left[x^2-\frac{x^4}{2}+\frac{x^6}{24}+....\right] \\ {\sin }^2\left(x\right)-x^2\cos \left(x\right)=\frac{x^4}{6}+\frac{x^6}{360}+..... \end{gathered}$$

Further, solve the limit $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\mathrm{x}}^2}-\frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\right)$as:

role="math" localid="1658556989686" $$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\mathrm{x}}^2}-\frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\right)=\frac{{\displaystyle \frac{{\mathrm{x}}^4}{6}+\frac{{\mathrm{x}}^6}{360}+...}}{{\mathrm{x}}^4-{\displaystyle \frac{{\mathrm{x}}^6}{3}}+{\displaystyle \frac{2{\mathrm{x}}^8}{45}}+...} \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\mathrm{x}}^2}-\frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\right)=\frac{{\displaystyle \frac{1}{6}+\frac{{\mathrm{x}}^2}{360}+....}}{1-{\displaystyle \frac{{\mathrm{x}}^2}{3}}+{\displaystyle \frac{2{\mathrm{x}}^4}{45}}+....} \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\mathrm{x}}^2}-\frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\right)=\frac{1}{6} \end{gathered}$$

Use the MATLAB tool to verify the limit as:

Hence, the required value of limits is mentioned as $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\mathrm{x}}^2}-\frac{\cos \left(\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right)}\right)=\frac{1}{6}$.

Rewrite the limit as,

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\left(\mathrm{x}\right)-\frac{1}{{\mathrm{x}}^2}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\sin }^2\left(\mathrm{x}\right)}-\frac{1}{{\mathrm{x}}^2}\right) \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{1}{{\sin }^2\left(\mathrm{x}\right)}-\frac{1}{{\mathrm{x}}^2}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(\frac{{\mathrm{x}}^2-{\sin }^2\left(\mathrm{x}\right)}{{\mathrm{x}}^2\sin \left(\mathrm{x}\right)}\right) \end{gathered}$$

Use Maclaurin series expansion of $\sin \left(x\right)$,

$$\begin{gathered} {\sin }^2\left(x\right)=x^2-\frac{x^4}{3}+\frac{2x^6}{45}+.... \\ {x}^2-{\sin }^2\left(x\right)=\frac{x^4}{3}-\frac{2x^6}{45}-.... \\ {x}^2{\sin }^2\left(x\right)=x^4-\frac{x^6}{3}+\frac{2x^8}{45}+.... \end{gathered}$$

Further, solve the limit $\underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right)$as:

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right)=\frac{{\displaystyle \frac{x^4}{3}}-{\displaystyle \frac{2x^6}{45}}-....}{x^4-{\displaystyle \frac{x^6}{3}}+{\displaystyle \frac{2x^8}{45}}+...} \\ \underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right)=\frac{{\displaystyle \frac{1}{3}}-{\displaystyle \frac{2x^2}{45}}-...}{1-{\displaystyle \frac{x^2}{3}}+{\displaystyle \frac{2x^4}{45}}} \\ \underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right)=\frac{1}{3} \end{gathered}$$

Use the MATLAB tool to verify the limit.

Hence, the required value of limits is mentioned as $\underset{\mathrm{x}\to 0}{\lim }\left({\csc }^2\mathrm{x}-\frac{1}{{\mathrm{x}}^2}\right)=\frac{1}{3}$.

Use Maclaurin series expansion of $\mathrm{In}\left(1+\mathrm{x}\right)$,

$$\begin{gathered} \mathrm{In}\left(1+\mathrm{x}\right)=\mathrm{x}-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}+.... \\ \mathrm{In}\left(1+\mathrm{x}\right)-\mathrm{x}=-\frac{{\mathrm{x}}^2}{2}+\frac{{\mathrm{x}}^3}{3}-\frac{{\mathrm{x}}^4}{4}+.... \\ \frac{\mathrm{In}\left(1+\mathrm{x}\right)-\mathrm{x}}{{\mathrm{x}}^2}=-\frac{1}{2}+\frac{\mathrm{x}}{3}-\frac{{\mathrm{x}}^2}{4}+.... \end{gathered}$$

Further, solve the limit $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{In}\left(1+\mathrm{x}\right)}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\right)$as:

$$\begin{gathered} \underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{In}\left(1+\mathrm{x}\right)}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\right)=\underset{\mathrm{x}\to 0}{\lim }\left(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+....\right) \\ \underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{In}\left(1+\mathrm{x}\right)}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\right)=-\frac{1}{2} \end{gathered}$$

Use the MATLAB tool to verify the limit.

Hence, the required value of limits is mentioned as $\underset{\mathrm{x}\to 0}{\lim }\left(\frac{\mathrm{In}\left(1+\mathrm{x}\right)}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}\right)=-\frac{1}{2}$.