How do you use L'hospital's rule to find the limit #lim_(x>0)(xsin(x))/(xtan(x))# ?
To use L'Hôpital's Rule to find the limit (\lim_{x \to 0} \frac{x  \sin(x)}{x  \tan(x)}), we can apply the rule to the fraction (\frac{0}{0}) form.
First, we differentiate the numerator and the denominator separately:

Differentiate the numerator: [ \frac{d}{dx}(x  \sin(x)) = 1  \cos(x) ]

Differentiate the denominator: [ \frac{d}{dx}(x  \tan(x)) = 1  \sec^2(x) ]
Next, we substitute these derivatives back into the original limit expression:
[ \lim_{x \to 0} \frac{1  \cos(x)}{1  \sec^2(x)} ]
Now, evaluate this new limit expression:

Substitute (x = 0) into the expression: [ \frac{1  \cos(0)}{1  \sec^2(0)} = \frac{1  1}{1  1} ]

Simplify the expression: [ \frac{0}{0} ]
Since we still have an indeterminate form, we can apply L'Hôpital's Rule again. Differentiate the numerator and denominator:

Differentiate the numerator: [ \frac{d}{dx}(1  \cos(x)) = \sin(x) ]

Differentiate the denominator: [ \frac{d}{dx}(1  \sec^2(x)) = 2\sec(x)\tan(x) ]
Substitute these derivatives back into the limit expression:
[ \lim_{x \to 0} \frac{\sin(x)}{2\sec(x)\tan(x)} ]
Now, evaluate the limit as (x) approaches 0:
 Substitute (x = 0) into the expression: [ \frac{\sin(0)}{2\sec(0)\tan(0)} = \frac{0}{0} ]
Since we still have an indeterminate form, we can apply L'Hôpital's Rule one more time. Differentiate the numerator and denominator:

Differentiate the numerator: [ \frac{d}{dx}(\sin(x)) = \cos(x) ]

Differentiate the denominator: [ \frac{d}{dx}(2\sec(x)\tan(x)) = 2\sec(x)\tan(x)  2\sec^3(x) ]
Substitute these derivatives back into the limit expression:
[ \lim_{x \to 0} \frac{\cos(x)}{2\sec(x)\tan(x)  2\sec^3(x)} ]
Finally, evaluate the limit as (x) approaches 0:
 Substitute (x = 0) into the expression: [ \frac{\cos(0)}{2\sec(0)\tan(0)  2\sec^3(0)} = \frac{1}{0  2} = \frac{1}{2} ]
So, (\lim_{x \to 0} \frac{x  \sin(x)}{x  \tan(x)} = \frac{1}{2}).
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The limit
Thus,
#lim_(x>0)(xsin(x))/(xtan(x)) = lim_(x>0){d/dx(xsin(x))}/ {d/dx(xtan(x))}#
We could use L'hospital's rule yet again, but it is much simpler to use simple trigonometry to go ahead :
and thus
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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