2 Answers
62tan x = x + x^3/3 + \text{higher powers} \\\frac{tan x}{x} = 1+x^2/3+\text{higher powers}
Now you have to find \\\lim_{x\rightarrow 0}\frac{tan x}{x} = \lim_{x\rightarrow 0}(1+x^2/3+\text{higher powers})^{3/x^2\times 3}
Which will be e^3
1SIR, I would like 2 add another way of solving it.
we know if \lim_{x\rightarrow 0}f(x) = 1 and \lim_{x\rightarrow 0}g(x) = infinity then
\lim_{x\rightarrow 0}[f(x)]^g^(^x^) = \lim_{x\rightarrow 0}[1 + {f(x)- 1}]^g^(^x^) = e\lim_{x\rightarrow 0}[{f(x)- 1}]g(x)
actually, here i have specifically taken a=0.
solving this way also, we cn get d same result.
