# AP Calculus AB/BC 1.5 Limits

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AP Calculus: AB/BC Limits Drill 1, Problem 5. Evaluate the limit.

AP Calculus | Limits |

Language | English Language |

MPAC 1 | Reasoning with definitions and theorems |

MPAC 3 | Implementing algebraic/computational processes |

### Transcript

On the top of the fraction, 11 plus 11 works out as 22...

...but when we plug it in to the denominator, we get that 11 squared is 121

and 121 minus 121 is 0.

Dividing by zero is a big no-no. You can't do that!

So to avoid that situation, let's try simplifying the bottom.

x squared minus 121 looks like a difference of two squares...

...a squared minus b squared equals a plus b times a minus b.

The square root of 121 is 11. So applying the formula, the bottom simplifies to x plus

11 times x minus 11.

The x plus 11 on the top and bottom of the fraction cancel out... and we're left with the

limit as x approaches 11 from the negative side of 1 over x minus 11.

Back to the question...it asks us to take the limit as x approaches 11 from the negative side.

Whenever we deal with limits, it means we get really close to a number from the negative

side but never actually kiss it.

If we try plugging in 11 as x again...11 minus 11 still gives us 0 in the denominator...

...so we can immediately eliminate (A), (B), and (C) because this function is not defined.

So we're left with D) or (E). We can now start plugging in values smaller

than 11 that get closer and closer to 11...since we're approaching 11 from the negative side.

If x equals 10, we get 1 over 10 minus 11, or negative 1.

If x equals 10.5... then 1 over negative .5 gives us

an even larger negative number...or negative 2.

Looks like we're getting more and more negative....and will keep approaching negative infinity...

So our answer is (E). Done!