Question Find the limit: \[ \lim _{t \rightarrow 0}\left\langle\frac{e^{-4 t}-1}{t}, \frac{t^{3}}{t^{4}-t^{3}}, \frac{6}{5+t}\right\rangle \]
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\lim _{t \rightarrow 0}\left\langle\frac{e^{-4 t}-1}{t}, \frac{t^{3}}{t^{4}-t^{3}}, \frac{6}{5+t}\right\rangle\rightarrow \lim _{t \rightarrow 0} \frac{e^{-4 t}-1}{t}=\frac{0}{0} \text { form }Ally 2-Hopita'' Rule:If \lim _{x \rightarrow a} \frac{f(x)}{s(x)}=\frac{0}{0} (or) \frac{\infty}{\infty} thes \lim _{x \rightarrow a} \frac{f(x)}{s(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{s^{\prime}(x)}\begin{array}{l}=\lim _{t \rightarrow 0} \frac{\frac{d}{d t}\left(e^{-4 t}-1\right)}{\frac{d}{d t}(t)}=\lim _{t \rightarrow 0} \frac{-4 e^{-4 t}-0}{1} \\=\frac{-4 e^{0}}{1}=-4\end{array}\begin{aligned}\rightarrow & \lim _{t \rightarrow 0} \frac{t^{3}}{t^{4}-t^{3}}=\lim _{t \rightarrow 0} \frac{t^{3}}{t^{3}(t-1)} \\& =\lim _{t \rightarrow 0} \frac{1}{t-1}=\frac{1}{0-1}=-1\end{aligned}\begin{array}{l} \rightarrow \lim _{t \rightarrow 0} \frac{6}{5+t}=\frac{6}{5+0}=6 / 5 \\=<-4,-1, \frac{6}{5}\end{array} ...