What is the limit as t approaches 5 of 1/(t^2-4t-5)-1/(6(t-5)) ?

The limit as t approaches 5 of 1/(t^2-4t-5)-1/(6(t-5)) is -1/36

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limit as t approaches 5 of 1/(t^2-4t-5)-1/(6(t-5))

t \to 5 lim (\frac{1}{t ^{2} - 4 t - 5} - \frac{1}{6 ( t - 5 )})

- \frac{1}{36}

Decimal

- 0.02777 \dots

Solution steps

t \to 5 lim (\frac{1}{t ^{2} - 4 t - 5} - \frac{1}{6 ( t - 5 )})

\frac{1}{t ^{2} - 4 t - 5} - \frac{1}{6 ( t - 5 )} = \frac{1}{t ^{2} - 4 t - 5} (1 - \frac{\frac{1}{6 ( t - 5 )}}{\frac{1}{t ^{2} - 4 t - 5}})

= t \to 5 lim (\frac{1}{t ^{2} - 4 t - 5} (1 - \frac{\frac{1}{6 ( t - 5 )}}{\frac{1}{t ^{2} - 4 t - 5}}))

Simplify

\frac{1}{t ^{2} - 4 t - 5} (1 - \frac{\frac{1}{6 ( t - 5 )}}{\frac{1}{t ^{2} - 4 t - 5}}) : - \frac{1}{6 ( t + 1 )}

= t \to 5 lim (- \frac{1}{6 ( t + 1 )})

Plug in the value

t = 5

= - \frac{1}{6 ( 5 + 1 )}

Simplify

= - \frac{1}{36}

What is the limit as t approaches 5 of 1/(t^2-4t-5)-1/(6(t-5)) ?
The limit as t approaches 5 of 1/(t^2-4t-5)-1/(6(t-5)) is -1/36