91Ó°ÊÓ

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

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,

$$\begin{gathered} e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+… \\ {e}^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+… \\ {e}^x-1-x=\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+… \\ x\left(e^x-1\right)=x^2+\frac{x^3}{2!}+\frac{x^4}{3!}+\frac{x^5}{4!}+… \end{gathered}$$

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

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

Use the MATLAB tool to verify the limit.

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

Use Maclaurin series expansion of sin (x),

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

Use Maclaurin series expansion of cos (x),

role="math" localid="1662641614300" $\begin{array}{rcl}\cos \left(x\right)& =& 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+…\\ x^2\cos \left(x\right)& =& x^2-\frac{x^4}{2}+\frac{x^6}{24}+…\\ {\sin }^2\left(x\right)-x^2\cos \left(x\right)& =& \left[x^2-\frac{x^4}{3}+\frac{2x^6}{45}+…\right]-\left[x^2-\frac{x^4}{2}+\frac{x^6}{24}+…\right]\\ {\sin }^2\left(x\right)-x^2\cos \left(x\right)& =& \frac{x^4}{6}+\frac{x^6}{360}+…\end{array}$

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

$$\begin{gathered} \underset{x→0}{\lim }\left(\frac{1}{x^2}-\frac{\cos \left(x\right)}{{\sin }^2\left(x\right)}\right)=\frac{\frac{x^4}{6}+\frac{x^6}{360}+…}{x^4-\frac{x^6}{3}+\frac{2x^8}{45}+…} \\ \underset{x→0}{\lim }\left(\frac{1}{x^2}-\frac{\cos \left(x\right)}{{\sin }^2\left(x\right)}\right)=\frac{\frac{1}{6}+\frac{x^2}{360}+…}{1-\frac{x^2}{3}+\frac{2x^4}{45}+…} \\ \underset{x→0}{\lim }\left(\frac{1}{x^2}-\frac{\cos \left(x\right)}{{\sin }^2\left(x\right)}\right)=\frac{1}{6} \end{gathered}$$

Use the MATLAB tool to verify the limit as:

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

Rewrite the limit as,

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

Use Maclaurin series expansion of sin (x),

$\begin{array}{rcl}{\sin }^2\left(x\right)& =& x^2-\frac{x^4}{3}+\frac{2x^6}{45}+…\\ x^2-{\sin }^2\left(x\right)& =& +\frac{x^4}{3}-\frac{2x^6}{45}-…\\ x^2{\sin }^2\left(x\right)& =& x^4-\frac{x^6}{3}+\frac{2x^8}{45}+…\end{array}$

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

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

Use the MATLAB tool to verify the limit.

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

Use Maclaurin series expansion of ln (1 + x),

$$\begin{gathered} \ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+… \\ \ln (1+x)-x=-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+… \\ \frac{\ln (1+x)-x}{x^2}=-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+… \end{gathered}$$

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

localid="1662642336152" $$\begin{gathered} \underset{x→0}{\lim }\left(\frac{\ln (1+x)}{x^2}-\frac{1}{x}\right)=\underset{x→0}{\lim }\left(-\frac{1}{2}+\frac{x}{3}-\frac{x^2}{4}+…\right) \\ \underset{x→0}{\lim }\left(\frac{\ln (1+x)}{x^2}-\frac{1}{x}\right)=-\frac{1}{2} \end{gathered}$$

Use the MATLAB tool to verify the limit.

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