How do you find the limit of #(3x+1/x) - (1/sinx)# as x approaches 0 using l'hospital's rule?

Answer 1

You can't evaluate the last expression using L'Hôpital's rule as it is not of an indeterminate form #0/0# or #oo/oo#

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Answer 2

This is a bit of a non-answer...

#(3x+1/x)-(1/sin x) = (3x^2 sin x + sin x - x) / (x sin x)#

Let:

#{ (f(x) = 3x^2 sin x + sin x - x), (g(x) = x sin x) :}#

Let:

#I = (-1, 1)#

Then:

#f(x)# and #g(x)# are differentiable on #I#
#lim_(x->0) f(x) = lim_(x->0) g(x) = 0#
#g'(x) = sin x + x cos x != 0# when #x in I "\" {0}#
#lim_(x->0) (f'(x))/(g'(x)) = lim_(x->0) (6x sin x + 3x^2 cos x + cos x - 1)/(sin x + x cos x)#
which is again in the form #0/0#
So it's not clear that #lim_(x->0) (f'(x))/(g'(x))# exists and the new form is more complicated than the original problem.

So L'Hôpital's rule does not help us much.

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Answer 3

To find the limit of the expression (3x+1/x) - (1/sinx) as x approaches 0 using L'Hospital's rule, we can differentiate the numerator and denominator separately and then evaluate the limit again.

First, let's differentiate the numerator and denominator:

The derivative of 3x+1 is 3. The derivative of 1/x is -1/x^2. The derivative of sinx is cosx.

Now, let's rewrite the expression with the derivatives:

(3) - (-1/x^2) / cosx

Next, let's simplify the expression:

3 + 1/x^2 / cosx

Now, let's evaluate the limit as x approaches 0:

As x approaches 0, 1/x^2 approaches infinity, and cosx approaches 1.

Therefore, the limit of the expression (3x+1/x) - (1/sinx) as x approaches 0 using L'Hospital's rule is:

3 + infinity / 1 = infinity.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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