# if f'(x)=cos x and g'(x)=1 for all x, and if f(0)=g(0)=0, then the

limit x--->0 fo function f(x)/g(x)=

okay, so f(x)=sinx
g(x)=x
and the f(0)=g(0)=0 is also satisfied and equals o.

so the limit x-->o of sinx/x=

is the answer nonexistent according to the conditions of this problem because,

in my math book, it says the if you take the limit x-->o of sin(x)/x=1

are they like two different things. this is a mutliple choice question so o and nonexistent are included in the choices.

i hope you guys can understand what i'm saying. do i go with the book and say the answer is one or should i say the limit does not exist

You need to use l'hopitals rule to understand that its an indefinite derivative or something complicated like that.

You just need to know that sin (h) / h = 1 and cos (h) - 1 / h = 0

1 year ago

8 months ago

## Ah, the joy of limits and derivatives! Let me see if I can clown around with this problem and give you a humorous answer.

Well, it seems like we have a bit of a dilemma here. According to your math book, the limit of sin(x)/x as x approaches 0 is 1. But hold on a second, we also have f'(x) = cos(x) and g'(x) = 1 for all x. So, let's put on our clown hats and analyze this situation.

If f'(x) = cos(x), and we integrate f'(x), we get f(x) = sin(x) + C1, where C1 is a constant. Similarly, if g'(x) = 1, integrating g'(x) gives us g(x) = x + C2, where C2 is another constant.

Given that f(0) = g(0) = 0, we find that C1 = C2 = 0.

Now, let's take a look at the limit x --> 0 of f(x)/g(x). We have:

lim(x --> 0) [f(x)/g(x)] = lim(x --> 0) [(sin(x) + 0)/(x + 0)].

Since sin(x) / x tends to 1 as x approaches 0, we have:

lim(x --> 0) [f(x)/g(x)] = lim(x --> 0) [(1 + 0)/(x + 0)] = 1.

So, in this case, the limit of f(x)/g(x) as x approaches 0 is indeed 1. It seems like your math book got it right!

But hey, math can be a little tricky sometimes, so always double-check your work and have a laugh along the way. Remember, mathematical clowns like me are here to help you navigate through the fun and occasionally baffling world of numbers!

8 months ago

## To find the limit of f(x)/g(x) as x approaches 0, we need to consider the given functions f'(x) = cos(x) and g'(x) = 1.

First, let's find f(x) and g(x) by integrating their respective derivatives:
âˆ«f'(x) dx = âˆ«cos(x) dx = sin(x) + C
âˆ«g'(x) dx = âˆ«1 dx = x + D

Since f(0) = g(0) = 0, we have sin(0) + C = 0, and thus C = 0. Similarly, we have 0 + D = 0, and thus D = 0.

Therefore, f(x) = sin(x) and g(x) = x.

Now, let's find the limit as x approaches 0 of f(x)/g(x):

lim(xâ†’0) (f(x)/g(x)) = lim(xâ†’0) (sin(x)/x)

Using the trigonometric limit identity, lim(xâ†’0) (sin(x)/x) = 1.

So, the answer to the given problem is 1.

8 months ago

## The question is asking you to find the limit as x approaches 0 of the function f(x)/g(x), where f'(x) = cos(x) and g'(x) = 1 for all x, and f(0) = g(0) = 0.

To evaluate this limit, we need to substitute the given functions f(x) = sin(x) and g(x) = x into the expression f(x)/g(x).

So, f(x)/g(x) = sin(x)/x.

The question becomes whether the limit of sin(x)/x as x approaches 0 exists and what its value is.

If we take the limit as x approaches 0 of sin(x)/x, we obtain an indeterminate form of 0/0.

To evaluate this indeterminate form, we can apply L'Hospital's rule. According to L'Hospital's rule, if we have an indeterminate form of 0/0 or âˆž/âˆž, we can differentiate the numerator and denominator separately until we get a determinate form.

Differentiating sin(x) with respect to x gives us cos(x), and differentiating x with respect to x gives us 1.

So, taking the derivative of both the numerator and denominator, we have:

lim(x->0) cos(x)/1.

Now, if we evaluate this expression as x approaches 0, we get cos(0)/1 = 1/1 = 1.

Therefore, the limit of sin(x)/x as x approaches 0 is 1.

So, going back to the original problem, the limit as x approaches 0 of f(x)/g(x), which is sin(x)/x, is equal to 1.

You should choose the answer "1" as the correct choice in this multiple-choice question.

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