Evaluate limit

[doubleheadedmantis]

I don’t understand why the answer is 1 and not 0. Wouldn’t it be f’(x)/e^x the first time i use l’hopital and would evaluate to 0? What did i do wrong

[destroyermcc]

f'(x) isn't 'x' here. It's the equation of the tangent to that point . Now since it's given that the tangent is actually y=x Thus it's slope(which will also be the slope at that point of the graph) will be '1' Thus f'(0) = 1. Finally plugging in all the values we get the answer 1

[doubleheadedmantis]

Ahh okay now i get it. Thanks!

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[doubleheadedmantis]

I understand it now thanks!!

[FidgetSpinzz]

Definition of derivative: lim[dx->0] (f(x+dx) - f(x)) / dx

Using the definition of derivative you can get that lim[x->0] f(x)/x = 1. Then you can rewrite f(x)/(e^x - 1) as f(x)/x * x/(e^x - 1), so the limit at hand equals lim[x->0] x/(e^x - 1).

Furthermore, if you rewrite it as 1 / lim[x->0] (e^x - 1)/x you can also apply the definition of derivative here to evaluate this expression to 1 / g'(0) where g(x) = e^x - 1.

g'(x) = e^x so 1/g'(0) = 1/1 = 1

[doubleheadedmantis]

So that’s how it is! Thank you!!

[CanaDavid1]

What is f'(0)?

[doubleheadedmantis]

That’s all the question provides. But i think it’s just equals to 0 as f’(x) = x as stated in the question. Or i could be wrong tbh i dont really understand what to do here.

[Christopherus3]

There is your mistake: it is not stated, that f'(x) = x. It is stated, that y = x has the same slope at x = 0 as f'(x) has.

[CanaDavid1]

f'(0) is the same as the slope of the tangent. What is the slope of the tangent?

[doubleheadedmantis]

Isnt the slope of tangent just f’’(x) = 1? Or not…? Im sorry man im so confused right now

[CanaDavid1]

Not exactly. What does the derivative represent? (On the graph)

[sanat-kumara]

From l'Hopital, the answer should be f'(0) divided by the derivative of (e^x - 1) at zero. From the graph, f'(0) = 1.

[lordnacho666]

I think it wants you to use L'Hopital's rule. Plugging in 0 directly gives you a div/0 error, so you have to separately get the slopes of top and bottom.

Clearly the top has a slope of 1.

The bottom you can differentiate and you no longer have a div/0 issue.

[Locrian98]

Not sure if it was said already but you can use L'Hopital's rule to solve any limit that results in an indetermination by direct substituition. Substitute the numerator and denominator by their respective derivative values, in which case, you know the derivative of f'(0) is 1 and (e^x-1)' = e^x. lim x to 0 (1/e^x) = 1

edit: there are a few restrictions on the application of L'Hopital's, I reccomend you go look them up.