## a non-trivial limit problem — calculus IIT JEE

Here is a what I call, a non-trivial problem of  evaluation of a limit. (Hey, I call it non-trivial because I could  only find a longish, brute-force solution as given below. If you have a clever, elegant solution, I would like to learn from you. Just write it down, take a snapshot in your smartphone camera, and post it in comments :-)).

Evaluate $\lim_{x \rightarrow 0} {((a^{x}+b^{x}+c^{3})/3)}^{2/x}$

Solution. The above limit can be recast as follows: $\lim_{x \rightarrow 0}{((a^{x}-1+b^{x}-1+c^{x}-1+3)/3)}^{2/x}$

= $\lim_{x \rightarrow 0}{((a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3+1)}^{2/x}$

Now, observe as $x \rightarrow 0$, $(a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3$ also tends to zero.

So, let $X=((a^{x}-1)/3+(b^{x}-1)/3+(c^{x}-1)/3)$. Hence, the required limit is equal to $\lim_{X \rightarrow 0}{(1+X)}^{(2/X)(X/x)}$

= $\lim_{X \rightarrow 0}{{((1+X)}^{(1/X)})}^{2X/x}$ call this A but we know that $lim_{X \rightarrow 0} {(1+X)}^{1/X}=e$

Hence, the above expression becomes $e^{(2/3)(( \lim_{x \rightarrow 0}(a^{x}-1)/x)+(\lim_{x \rightarrow 0}(b^{x}-1)/x)+(\lim_{x \rightarrow 0}(c^{x}-1)/x))}$

= $e^{(2/3)((\ln a)+(\ln b)+(\ln c))}$

= $e^{(2/3)(\ln (abc))}$

= ${(abc)}^{2/3}$.

So, what do you think — is there an alternate elegant solution to  my brute force solution above? if you have, please share it with me also.

More later…

Nalin Pithwa

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