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  • Choose the right expansion which is represented by the limit  <img alt="\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim%20_%7Bx%20%5Crigh

Choose the right expansion which is represented by the limit  \lim _{x \rightarrow 0} \frac{a^{x}-1}{x}.

Option: 1

\lim _{x \rightarrow 0}\left(\frac{x^{2} \ln e}{1 !}+\frac{x^{3}(\ln e)^{2}}{2 !}+..........\right)


Option: 2

\lim _{x \rightarrow 0}\left(\frac{\ln e}{1 !}+\frac{x(\ln e)^{2}}{2 !}+\frac{x^{2}(\ln e)^{3}}{3 !}+\right)


Option: 3

\lim _{x \rightarrow 0}\left(\frac{(\ln e)}{1 !}+\frac{x^{2}(\ln e)}{2 !}+\frac{x^{3}(\ln e)}{3 !}+ ............\right)


Option: 4

\lim _{x \rightarrow 0}\left(\frac{x(\ln e)}{1 !}+\frac{x^{2}(\ln e)}{2 !}+..................\right)
 


Answers (1)

best_answer

The Taylor expansion of e^{x} is:

e^{x}=1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+............

Now, subtract 1 from both the sides of the equation (i).

\begin{aligned} e^{x}-1 & =\left(1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+...........\right)-1 \\ e^{x}-1 & =\left(\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+............\right) \end{aligned}

As x \neq 0, divide both the sides of the equation (ii) by x.

\frac{e^{x}-1}{x}=\frac{\left(\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+...........\right)}{x}

\frac{e^{x}-1}{x}=\left(\frac{1}{1 !}+\frac{x}{2 !}+\frac{x^{2}}{3 !}+\right)

It is evident that using the equation (iii), the following can be written.

\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \\ & =\lim _{x \rightarrow 0}\left(1+\frac{x}{2 !}+\frac{x^{2}}{3 !}+.............\right) \\ & =\lim _{x \rightarrow 0}\left(\frac{\ln e}{1 !}+\frac{x(\ln e)^{2}}{2 !}+\frac{x^{2}(\ln e)^{3}}{3 !}+\right) \quad\left[\bigvee \ln e=\log _{e} e=1\right] \\ & =1 \end{aligned}

Posted by

sudhir.kumar

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