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If $f(t)=\int_{-t}^t \frac{e^{-|x|}}{2} d x$, then $\lim _{t \rightarrow \infty} f(t)$ is equal
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Verified Answer
The correct answer is:
$1$
$\begin{aligned} f(t) & =\int_{-t}^t \frac{e^{-|x|}}{2} d x \\ & =2 \int_0^t \frac{e^{-x}}{2} d x \\ & =-\left[e^{-x}\right]_0^t=-e^{-t}+1\end{aligned}$
Now, $\begin{aligned} \lim _{t \rightarrow \infty} f(t) & =-\lim _{t \rightarrow \infty} e^{-t}+1 \\ & =0+1 \\ & =1\end{aligned}$
Now, $\begin{aligned} \lim _{t \rightarrow \infty} f(t) & =-\lim _{t \rightarrow \infty} e^{-t}+1 \\ & =0+1 \\ & =1\end{aligned}$
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