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

Find lim_(π‘₯ β†’ 0) ((19^(π‘₯)) βˆ’ 1)/(√(π‘₯ + 25) βˆ’ 5).

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Video Transcript

Find the limit as π‘₯ approaches zero of 19 to the power of π‘₯ minus one over the square root of π‘₯ plus 25 minus five.

If we try to evaluate this limit by finding the limit as π‘₯ approaches zero of 19 to the power of π‘₯ minus one over the limit as π‘₯ approaches zero of the square root of π‘₯ plus 25 minus five by direct substitution of π‘₯ equals zero, we would end up with one minus one over five minus five. Because, recall, that any constant raised to the power of zero is one. So we end up with zero over zero, an indeterminate form, which is not very helpful to us. And so we need to evaluate this limit using L’HΓ΄pital’s rule.

L’HΓ΄pital’s rule says that if the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ equals zero and the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ equals zero or the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ is positive or negative infinity and the limit as π‘₯ approaches π‘Ž of 𝑔 of π‘₯ is positive or negative infinity, then the limit as π‘₯ approaches π‘Ž of 𝑓 of π‘₯ over 𝑔 of π‘₯ is the limit as π‘₯ approaches π‘Ž of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯.

Remember, we got zero over zero when we tried to evaluate this limit. So, we’re in this first scenario. Because 19 to the power of π‘₯ minus one is our 𝑓 of π‘₯, and the square root of π‘₯ plus 25 minus five is our 𝑔 of π‘₯, so we’re going to apply L’HΓ΄pital’s rule. And I’m gonna clear some space on the screen. But I’m just going to keep this bit, as this is what we’re going to need.

Okay, so we said that 𝑓 of π‘₯ is 19 to the power of π‘₯ minus one. And we want to find 𝑓 prime of π‘₯. So what we need to remember here is that if 𝑦 equals π‘Ž to the power of π‘₯, then d𝑦 by dπ‘₯ equals π‘Ž to the power of π‘₯ ln π‘Ž. And so 𝑓 prime of π‘₯ equals 19 to the power of π‘₯ ln 19. Because, remember, one is a constant, so it differentiates to zero. So, that’s 𝑓 prime of π‘₯. Remember 𝑔 of π‘₯ is the square root of π‘₯ plus 25 minus five. And here we have a function of a function. So, we’re going to need the chain rule.

I’m just going to rewrite this slightly because, remember, taking the square root of something is the same as raising it to the power of a half. So, this is π‘₯ plus 25 to the power of a half minus five. And, remember, that the chain rule says that if 𝑦 equals 𝑓 of 𝑒 and 𝑒 equals 𝑔 of π‘₯, then d𝑦 by dπ‘₯ equals d𝑦 by d𝑒 multiplied by d𝑒 by dπ‘₯.

So, let’s go ahead and differentiate π‘₯ plus 25 to the power of a half using the chain rule. If we start by letting 𝑦 equal π‘₯ plus 25 to the power of a half. So, 𝑒 is going to be π‘₯ plus 25. And differentiating that, with respect to π‘₯, gives us that d𝑒 by dπ‘₯ equals one. And because we let 𝑒 equal π‘₯ plus 25, then we can say that 𝑦 equals 𝑒 to the power of a half. And by multiplying by the power and then subtracting one from the power with usual differentiation rules that we know, we find that d𝑦 by d𝑒 equals a half 𝑒 to the power of negative a half.

And by the formula for the chain rule, we have that d𝑦 by dπ‘₯ equals a half 𝑒 to the power of negative a half multiplied by one, which is just a half 𝑒 to the power of negative a half. And, remember, we said that 𝑒 is equal to π‘₯ plus 25. So, we can replace 𝑒 to get a half π‘₯ plus 25 to the power of negative a half. And, remember, that raising something to the power of negative a half is the same as doing one over the square root of it. And from here, we can just tidy this up a little bit more by multiplying these two fractions together, to give us one over two multiplied by the square root of π‘₯ plus 25. And that gives us our 𝑔 prime of π‘₯.

So, now that we have our 𝑓 prime of π‘₯ and our 𝑔 prime of π‘₯, we can apply L’HΓ΄pital’s rule. So, I’ll clear some space for us to do that. So, from L’HΓ΄pital’s rule, we have that the limit, as π‘₯ approaches zero, of 19 to the power of π‘₯ minus one over the square root of π‘₯ plus 25 minus five is equal to the limit, as π‘₯ approaches zero, of 19 to the power of π‘₯ ln 19 over one over two multiplied by the square root of π‘₯ plus 25.

And we know from our rules of dividing fractions that this is the same as the limit, as π‘₯ approaches zero, of 19 to the power of π‘₯ ln 19 multiplied by two multiplied by the square root of π‘₯ plus 25, all over one. Which is just the limit, as π‘₯ approaches zero, of 19 to the power of π‘₯ ln 19 multiplied by two multiplied by the square root of π‘₯ plus 25. And we can evaluate this limit by directly substituting π‘₯ equals zero. We remember that any constant to the power of zero is just one. And we know that the square root of 25 is five. So, finally, we end up with ln 19 times 10, which we usually write as 10 ln 19. So, that’s our final answer.

Finally, let’s just check that we met the conditions for L’HΓ΄pital’s rule. 𝑓 and 𝑔 are differentiable because we managed to differentiate them. Around the limit point, π‘₯ is zero, 𝑔 prime of π‘₯ is not equal to zero. And, finally, the limit, as π‘₯ approaches π‘Ž, of 𝑓 prime of π‘₯ over 𝑔 prime of π‘₯ exists because we managed to find it. So, all conditions were satisfied.

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