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Find the limit as π₯ approaches two of nine to the power of π₯ minus 81 over seven to the power of π₯ minus 49.
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If we try to evaluate this limit by evaluating the limit as π₯ approaches two of nine to the power of π₯ minus 81 over the limit as π₯ approaches two of seven to the power of π₯ minus 49, we would get nine squared minus 81 over seven squared minus 49.
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And this is 81 minus 81 over 49 minus 49.
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And this is just zero over zero, which is an indeterminate form.
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So, instead, we evaluate this limit by using lβHopitalβs rule.
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LβHopitalβs rule states that if the limit as π₯ approaches π of π of π₯ equals zero and the limit as π₯ approaches π of π of π₯ equals zero.
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Or the limit as π₯ approaches π of π of π₯ equals positive or negative β and the limit as π₯ approaches π of π of π₯ is positive or negative β.
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Then the limit as π₯ approaches π of π of π₯ over π of π₯ is equal to the limit as π₯ approaches π of π prime of π₯ over π prime of π₯.
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Remember, when we initially tried to evaluate our limit, we got zero over zero by direct substitution.
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So we have this first scenario.
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So letβs apply lβHopitalβs rule to our question.
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We can see that, for our problem, π of π₯ is nine to the power of π₯ minus 81.
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And π of π₯ is seven to the power of π₯ minus 49.
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So letβs go ahead and differentiate the top function.
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So whatβs the derivative of nine to the power of π₯ minus 81?
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Well, one helpful rule that weβre going to need here is that if π¦ equals π to the power of π₯, so a constant raised to the power of π₯, then dπ¦ by dπ₯ equals π to the power of π₯ multiplied by ln π₯.
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So nine to the power of π₯ minus 81 will differentiate to nine to the power of π₯ multiplied by ln nine.
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As 81 is a constant, it differentiates to zero.
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And for the same reasons, seven to the power of π₯ minus 49 differentiates to seven to the power of π₯ multiplied by ln seven.
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And so, by lβHopitalβs rule, the limit as π₯ approaches two of nine to the power of π₯ minus 81 over seven to the power of π₯ minus 49 is equal to the limit as π₯ approaches two of nine to the power of π₯ multiplied by ln nine over seven to the power of π₯ multiplied by ln seven.
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And if we now evaluate this limit by direct substitution of π₯ equals two, we get nine squared multiplied by ln nine over seven squared multiplied by ln seven, which is just 81 multiplied by ln nine over 49 multiplied by ln seven.
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Now, itβs important to know some conditions for lβHopitalβs rule.
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Firstly π and π must both be differentiable, which they are, as we differentiated nine to the power of π₯ minus 81 to get nine to the power of π₯ multiplied by ln nine.
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And we differentiated seven to the power of π₯ minus 49 to get seven to the power of π₯ multiplied by ln seven.
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So that condition is satisfied.
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Secondly, we must have that π prime of π₯ is not equal to zero near π.
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And for our question, π prime of π₯ was seven to the power of π₯ multiplied by ln seven, which is not zero near the limit 0.2.
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So that condition is satisfied.
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And finally the limit as π₯ approaches π of π prime of π₯ over π prime of π₯ must exist or be positive or negative β.
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And for our question, this does exist, because weβre able to find it.
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So the final condition was satisfied.