Question

Please give solution

answer is given 0
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Answer 1

On taking $t=\infty$ directly we get the indeterminate form $\dfrac{\infty}{\infty}.$

Hence by L'hospital's Rule,

$\displaystyle\lim_{t\to\infty}\dfrac{e^{t^2}-1}{e^{t^2}+1}=\lim_{t\to\infty}\dfrac{\frac{d}{dt}\left(e^{t^2}\right)-\dfrac{d}{dt}(1)}{\frac{d}{dt}\left(e^{t^2}\right)+\dfrac{d}{dt}(1)}\\\\\\\lim_{t\to\infty}\dfrac{e^{t^2}-1}{e^{t^2}+1}=\lim_{t\to\infty}\dfrac{\frac{d}{dt}\left(e^{t^2}\right)}{\frac{d}{dt}\left(e^{t^2}\right)}\\\\\\\lim_{t\to\infty}\dfrac{e^{t^2}-1}{e^{t^2}+1}=\lim_{t\to\infty}1\\\\\\\lim_{t\to\infty}\dfrac{e^{t^2}-1}{e^{t^2}+1}=\underline{\underline{\mathbf{1}}}$

Hence 1 is the answer.