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  4. The value of displaystyle lim x arrow ∞ ((2xn)(1/ex)-(3xn)(1/ex)/xn)( where n ∈ N) is

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Q. The value of $\displaystyle\lim _{x \rightarrow \infty} \frac{\left(2^{x^{n}}\right)^{\frac{1}{e^{x}}}-\left(3^{x^{n}}\right)^{\frac{1}{e^{x}}}}{x^{n}}($ where $n \in N)$ is

Limits and Derivatives

Solution:

$L=\displaystyle\lim _{x \rightarrow \infty} \frac{\left(2^{x^{n}}\right)^{\frac{1}{e^{x}}}-\left(3^{x^{n}}\right)^{\frac{1}{e^{x}}}}{x^{n}}$
$=\displaystyle\lim _{x \rightarrow \infty} \frac{(3)^{\frac{x^{n}}{e^{x}}}\left(\left(\frac{2}{3}\right)^{\frac{x^{n}}{e^{x}}}-1\right)}{x^{n}}$
Now, $\displaystyle\lim _{x \rightarrow \infty} \frac{x^{n}}{e^{x}}=\displaystyle\lim _{x \rightarrow \infty} \frac{n !}{e^{x}}=0 $
(Differentiating numerator and denominator $n$ times for L'Hospital's rule)
Hence, $L=\displaystyle\lim _{x \rightarrow \infty}(3)^{\frac{x^{n}}{x^{x}}} \displaystyle\lim _{x \rightarrow \infty} \frac{\left(\left(\frac{2}{3}\right)^{\frac{x^{n}}{e^{x}}}-1\right)}{\frac{x^{n}}{e^{x}}} \displaystyle\lim _{x \rightarrow \infty} \frac{1}{e^{x}}$
$=1 \times \log (2 / 3) \times 0=0$