Question

Lim x tende 0 ((2^t)-1)/t?

Answer 1

Calcular o limite

$L=\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{2^{t}-1}{t}$


No limite acima, temos uma indeterminação do tipo $0/0.$ Logo, podemos usar a regra de L'Hôpital para resolver:


Se o limite $\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{\frac{d}{dt}(2^{t}-1)}{\frac{d}{dt}(t)}$ existir (finito ou infinito), então

$\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{2^{t}-1}{t}=\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{\frac{d}{dt}(2^{t}-1)}{\frac{d}{dt}(t)}$


$\bullet\;\;$ Vamos calcular o limite pela regra de L'Hôpital:

$\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{\frac{d}{dt}(2^{t}-1)}{\frac{d}{dt}(t)}\\ \\ \\ =\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{2^{t}\,\mathrm{\ell n\,}{2}}{1}\\ \\ \\ =2^{0}\,\mathrm{\ell n\,}{2}\\ \\ =\mathrm{\ell n\,}{2}$


$\bullet\;\;$ Portanto,

$L=\underset{t \to 0}{\mathrm{\ell im}}\,\dfrac{2^{t}-1}{t}=\mathrm{\ell n\,}{2}$