If limx→$$0$$ $$1+x+fxx1/x=e3$$, then the value of lnlimx→$$0$$ $$1+fxx1/x$$ is $$______$$

limx→$$0$$ $$1+x+fxx1/x=e3or$$ elimx→$$0$$ $$1+x+fxx-11x=e3or$$ elimx→$$0$$ $$1+fxx2=e3$$ or limx→$$0$$ $$fxx2=2Now$$, limx→$$0$$ $$1+fxx1/x=elimx$$→$$0$$ $$1+fxx-11x=elimx$$→$$0$$ $$fxx2=e2$$

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$If \lim_{x \to 0} {[1 + x + \frac{f (x)}{x}]}^{1 / x} = e^{3}, then the value of \ln (\lim_{x \to 0} {[1 + \frac{f (x)}{x}]}^{1 / x}) is$ ______

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$\begin{array}{l} \lim_{x \to 0} {[1 + x + \frac{f (x)}{x}]}^{1 / x} = e^{3} \\ or e^{\lim_{x \to 0} [1 + x + \frac{f (x)}{x} - 1] \frac{1}{x}} = e^{3} \\ or e^{\lim_{x \to 0} [1 + \frac{f (x)}{x^{2}}]} = e^{3} or \lim_{x \to 0} \frac{f (x)}{x^{2}} = 2 \\ Now, \lim_{x \to 0} {[1 + \frac{f (x)}{x}]}^{1 / x} = e^{\lim_{x \to 0} [1 + \frac{f (x)}{x} - 1] \frac{1}{x}} = e^{\lim_{x \to 0} \frac{f (x)}{x^{2}}} = e^{2} \end{array}$

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Introduction to limits

Remember concepts with our Masterclasses.

If limx→0 1+x+fxx1/x=e3, then the value of lnlimx→0 1+fxx1/x is ______

limx→0 1+x+fxx1/x=e3or elimx→0 1+x+fxx-11x=e3or elimx→0 1+fxx2=e3 or limx→0 fxx2=2Now, limx→0 1+fxx1/x=elimx→0 1+fxx-11x=elimx→0 fxx2=e2

Ready to Test Your Skills?

free study material

$If \lim_{x \to 0} {[1 + x + \frac{f (x)}{x}]}^{1 / x} = e^{3}, then the value of \ln (\lim_{x \to 0} {[1 + \frac{f (x)}{x}]}^{1 / x}) is$ ______