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

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

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

Let:

Let:

Then:

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

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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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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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