Question

the given function is
${(1 - \frac{1}{3x} )}^{ {x}^{2} }$
and the limit x tends to infinity
.
.
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best answer = mark as brainliest​

Answer 1

Answer:

Step-by-step explanation:

the given function is $\lim_{x\to\infty} (1 - \frac{1}{3x})^{x^2}$

The answer is   $\lim_{x\to\infty} (1 - \frac{1}{3x})^{x^2} = 0$

$\lim_{x\to\infty} (1 - \frac{1}{3x})^{x^2} \hspace{0.1cm}=$    $\hspace{0.1cm} \lim_{x\to\infty} e^{\ln( (1 - \frac{1}{3x})^{x^2}) \hspace{0.1cm} = \hspace{0.1cm} \lim_{x\to\infty} e^{x^{2} \ln( (1 - \frac{1}{3x}))$

$= \hspace{0.1cm} \lim_{x\to\infty} e^{x^{2} \ln( (1 - \frac{1}{3x})) = \hspace{0.1cm} e^{\lim_{x\to\infty} x^{2} \ln( (1 - \frac{1}{3x}))$

$= \hspace{0.1cm} e^{\lim_{x\to\infty}\frac{1}{\frac{d}{dx}(\frac{1}{ x^{2}} )} \frac{d}{dx} \ln( (1 - \frac{1}{3x}))$

$= \hspace{0.1cm} e^{-\frac{1}{2}\frac{\lim_{x\to\infty x}}{\lim_{x\to\infty}( 3 - \frac{1}{x} )}$ = $e^{\frac{1}{2} (\frac{-\infty}{3})} = e^{-\infty} = 0$