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.
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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