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CENBVAL Mavegador Sers re-y-continuidad-de-fundiones/x7193e35109289adercalculo-d Khan Academy Límites por sustitución direct: Google Classroom Puede gue necesite Calculadora Encuentra lim_(x rarr3)f(x) para f(x)=(x^(2)-4)/(11-2x). Mostrar calculadora

CENBVAL Mavegador Sers\newlinere-y-continuidad-de-fundiones/x71937193e3510928935109289adercalculo-d\newlineKhan Academy\newlineLímites por sustitución direct:\newlineGoogle Classroom\newlinePuede gue necesite\newlineCalculadora\newlineEncuentra lim⁡x→3f(x) \lim _{x \rightarrow 3} f(x) para f(x)=x2−411−2x f(x)=\frac{x^{2}-4}{11-2 x} .\newlineMostrar calculadora

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Q. CENBVAL Mavegador Sers\newlinere-y-continuidad-de-fundiones/x71937193e3510928935109289adercalculo-d\newlineKhan Academy\newlineLímites por sustitución direct:\newlineGoogle Classroom\newlinePuede gue necesite\newlineCalculadora\newlineEncuentra lim⁡x→3f(x) \lim _{x \rightarrow 3} f(x) para f(x)=x2−411−2x f(x)=\frac{x^{2}-4}{11-2 x} .\newlineMostrar calculadora
  1. Identify Function & Point: Identify the function and the point of interest. f(x)=x2−411−2xf(x) = \frac{x^2 - 4}{11 - 2x}, we need to find lim⁡x→3f(x)\lim_{x \to 3} f(x).
  2. Substitute x=3x = 3: Substitute x=3x = 3 directly into the function to check if the function is defined at this point.f(3)=(32)−411−2⋅3=9−411−6=55=1.f(3) = \frac{(3^2) - 4}{11 - 2 \cdot 3} = \frac{9 - 4}{11 - 6} = \frac{5}{5} = 1.
  3. Calculate Limit: Since direct substitution did not result in an indeterminate form, the limit as xx approaches 33 is simply the value of the function at x=3x = 3.lim⁡x→3f(x)=1.\lim_{x \to 3} f(x) = 1.

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