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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%5Crightarrow

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{\log a}{1 !}+\frac{x(\log a)^{2}}{2 !}+........\right)


Option: 2

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


Option: 3

\lim _{x \rightarrow 0}\left(\frac{(\log a)}{1 !}+\frac{x(\log a)}{2 !}+........\right)


Option: 4

\lim _{x \rightarrow 0}\left(\frac{x(\log a)}{1 !}+\frac{x^{2}(\log a)}{2 !}+\right)


Answers (1)

best_answer

The Taylor expansion of a^{x} is:

a^{x}=\left(1+\frac{x(\log a)}{1 !}+\frac{x^{2}(\log a)^{2}}{2 !}+\frac{x^{3}(\log a)^{3}}{3 !}+...........\right).........(i)

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

\begin{aligned} & a^{x}-1=\left(1+\frac{x(\log a)}{1 !}+\frac{x^{2}(\log a)^{2}}{2 !}+\frac{x^{3}(\log a)^{3}}{3 !}+ .........\right)-1 \\ & a^{x}-1=\left(\frac{x(\log a)}{1 !}+\frac{x^{2}(\log a)^{2}}{2 !}+\frac{x^{3}(\log a)^{3}}{3 !}+........\right) \quad \ldots(ii) \end{aligned}

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

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

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

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

 

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SANGALDEEP SINGH

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