6.5: Regla de L'Hopital
- Page ID
- 108884
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)El siguiente resultado es un caso de regla de\(l^{\prime}\) L'hópital.
Supongamos\(a, b \in \mathbb{R}, f\) y\(g\) son diferenciables\((a, b), g^{\prime}(x) \neq 0\) para todos\(x \in(a, b),\) y
\[\lim _{x \rightarrow a^{+}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda .\]
Si\(\lim _{x \rightarrow a^{+}} f(x)=0\) y\(\lim _{x \rightarrow a^{+}} g(x)=0,\) entonces
\[\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .\]
- Prueba
-
Dado\(\epsilon>0,\) que existe\(\delta>0\) tal que
\[\lambda-\frac{\epsilon}{2}<\frac{f^{\prime}(x)}{g^{\prime}(x)}<\lambda+\frac{\epsilon}{2}\]
siempre\(x \in(a, a+\delta) .\) Ahora, por el Teorema del Valor Medio Generalizado, para cualquiera\(x\) y\(y\) con\(a<x<y<a+\delta,\) existe un punto\(c \in(x, y)\) tal que
\[\frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f^{\prime}(c)}{g^{\prime}(c)}.\]
De ahí
\[\lambda-\frac{\epsilon}{2}<\frac{f(y)-f(x)}{g(y)-g(x)}<\lambda+\frac{\epsilon}{2}.\]
Ahora
\[\lim _{x \rightarrow a^{+}} \frac{f(y)-f(x)}{g(y)-g(x)}=\frac{f(y)}{g(y)}\]
y así tenemos
\[\lambda-\epsilon<\lambda-\frac{\epsilon}{2} \leq \frac{f(y)}{g(y)} \leq \lambda+\frac{\epsilon}{2}<\lambda+\epsilon\]
para cualquier\(y \in(a, a+\delta) .\) Por lo tanto
\[\lim _{x \rightarrow a^{+}} \frac{f(x)}{g(x)}=\lambda .\]
Q.E.D.
Usa la regla de L'hôpital para calcular
\[\lim _{x \rightarrow 0^{+}} \frac{\sqrt{1+x}-1}{x}. \nonumber\]
Supongamos\(a, b \in \mathbb{R}, f\) y\(g\) son diferenciables\((a, b), g^{\prime}(x) \neq 0\) para todos\(x \in(a, b),\) y
\[\lim _{x \rightarrow b^{-}} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lambda . \nonumber\]
Demostrar que si\(\lim _{x \rightarrow b^{-}} f(x)=0\) y\(\lim _{x \rightarrow b^{-}} g(x)=0,\) entonces
\[\lim _{x \rightarrow b^{-}} \frac{f(x)}{g(x)}=\lambda . \nonumber\]