Video: Finding the Value of a Limit Involving Exponential Functions Using L’Hopital’s Rule

Find lim_(π‘₯ β†’ 0) (7𝑒^(5π‘₯) βˆ’ 7)/(βˆ’π‘’^(8π‘₯) + 1).

03:09

Video Transcript

Find the limit as π‘₯ approaches zero of seven 𝑒 to the five π‘₯ minus seven over negative 𝑒 to the eight π‘₯ plus one.

We will start by trying to solve this limit using direct substitution. We obtain seven times 𝑒 to the power of five times zero minus seven over negative 𝑒 to the power of eight times zero plus one. Since 𝑒 to the power of zero is equal to one, we find that this is equal to seven minus seven over negative one plus one, which simplifies to zero over zero. However, this is undefined. Although we obtain that our limit is undefined using direct substitution, it is equal to zero over zero. And this tells us that we may be able to use L’Hopital’s rule.

L’Hopital’s rule tells us that if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to zero over zero, positive infinity over positive infinity, or negative infinity over negative infinity. Where π‘Ž is a real number, positive infinity, or negative infinity. Then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is equal to the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯.

Now our limit satisfies the condition of our limit being equal to zero over zero. And since we’re taking the limit as π‘₯ approaches zero, that means that our π‘Ž is equal to zero, which is a real number. Therefore, we can use L’Hopital’s rule. 𝑓 of π‘₯ is the numerator of the function which we’re taking the limit of. So that’s seven 𝑒 to the five π‘₯ minus seven. And 𝑔 of π‘₯ is the denominator. So that’s negative 𝑒 to the power of eight π‘₯ plus one.

Now we must find 𝑓 prime of π‘₯ and 𝑔 prime of π‘₯. Since we will be differentiating exponential terms, we can use the rule that tells us that the differential of 𝑒 to the power of π‘˜π‘₯ with respect to π‘₯ is equal to π‘˜ times 𝑒 to the power of π‘˜π‘₯. Let’s differentiate 𝑓 of π‘₯ term by term. Seven 𝑒 to the power of five π‘₯ is an exponential term. So we’ll be using the rule which we’ve just stated. Our value of π‘˜ is five. And we notice that we have a constant of seven multiplying our exponential term. So that must remain two, giving us seven timesed by five 𝑒 to the five π‘₯. Then seven times five is 35. So we can write this as 35𝑒 to the power of five π‘₯.

The second term in 𝑓 of π‘₯ is negative seven, which is simply a constant. And when we differentiate any constant, we simply get zero. So we found that 𝑓 prime of π‘₯ is equal to 35𝑒 to the power of five π‘₯. The first term in 𝑔 of π‘₯ is negative 𝑒 to the power of eight π‘₯, which is again an exponential term. Using our rule, we obtain that the differential of this term is negative eight 𝑒 to the power of eight π‘₯. The second term in 𝑔 of π‘₯ is one, which is again a constant. And so this will differentiate to give zero.

We are now ready to apply L’Hopital’s rule. We obtain that the limit as π‘₯ approaches zero of seven 𝑒 to the power of five π‘₯ minus seven over negative 𝑒 to the power of eight π‘₯ plus one. Is equal to the limit as π‘₯ approaches zero of 35 times 𝑒 to the power of five π‘₯ over negative eight times 𝑒 to the power of eight π‘₯. And we can now apply direct substitution, giving us 35 timesed by 𝑒 to the power of zero over negative eight timesed by 𝑒 to the power of zero. Since 𝑒 to the power of zero is equal to one, we get a solution that our limit must be equal to negative 35 over eight.

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