find \\( \\lim _{t \\rightarrow 7} \\frac{t^{2}+t - 56}{t^{2}-49} \\).\n\nselect the correct choice below…

find \\( \\lim _{t \\rightarrow 7} \\frac{t^{2}+t - 56}{t^{2}-49} \\).\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. \\( \\lim _{t \\rightarrow 7} \\frac{t^{2}+t - 56}{t^{2}-49}= \\) (type an integer or a simplified fraction.)\n\nb. the limit does not exist.

find \\( \\lim _{t \\rightarrow 7} \\frac{t^{2}+t - 56}{t^{2}-49} \\).\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. \\( \\lim _{t \\rightarrow 7} \\frac{t^{2}+t - 56}{t^{2}-49}= \\) (type an integer or a simplified fraction.)\n\nb. the limit does not exist.

Answer

Explanation:

Step1: Factor the numerator and denominator

  • Factor the numerator (t^{2}+t - 56): Using the formula (ax^{2}+bx + c=a(x - x_1)(x - x_2)), where (x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for (ax^{2}+bx + c = 0). For (t^{2}+t - 56), (a = 1), (b = 1), (c=-56). Then (t=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-56)}}{2\times1}=\frac{-1\pm\sqrt{1 + 224}}{2}=\frac{-1\pm15}{2}). The roots are (t_1 = 7) and (t_2=-8), so (t^{2}+t - 56=(t - 7)(t + 8)).
  • Factor the denominator (t^{2}-49): Using the difference - of - squares formula (a^{2}-b^{2}=(a + b)(a - b)), where (a=t) and (b = 7). So (t^{2}-49=(t - 7)(t + 7)).

The limit becomes (\lim_{t\rightarrow7}\frac{t^{2}+t - 56}{t^{2}-49}=\lim_{t\rightarrow7}\frac{(t - 7)(t + 8)}{(t - 7)(t + 7)}).

Step2: Simplify the function (cancel out the common factor)

Since (t\neq7) when taking the limit as (t\rightarrow7) (we are approaching (t = 7) but not equal to (7)), we can cancel out the common factor ((t - 7)). The simplified function is (\lim_{t\rightarrow7}\frac{t + 8}{t + 7}).

Step3: Substitute (t = 7) into the simplified function

Substitute (t=7) into (\frac{t + 8}{t + 7}), we get (\frac{7+8}{7+7}=\frac{15}{14}).

Answer:

A. (\lim_{t\rightarrow7}\frac{t^{2}+t - 56}{t^{2}-49}=\frac{15}{14})