How to solve without the l'Hospital's rule? #lim_(x>0) (xcos^2(x))/(x+tan(3x))#
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Please see below.
So in the limit, we have:
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To evaluate the limit ( \lim_{x \to 0} \frac{x \cos^2(x)}{x + \tan(3x)} ) without using L'Hôpital's rule, we can simplify the expression by factoring and canceling common terms. Let's proceed step by step:

Substitute ( x = 0 ) into the expression: [ \lim_{x \to 0} \frac{x \cos^2(x)}{x + \tan(3x)} = \frac{0 \cdot \cos^2(0)}{0 + \tan(0)} ]

Simplify the expression with known trigonometric values: [ \frac{0 \cdot 1}{0 + 0} = \frac{0}{0} ]

Since we have an indeterminate form of ( \frac{0}{0} ), we can try simplifying further by factoring common terms: [ \lim_{x \to 0} \frac{x \cos^2(x)}{x + \tan(3x)} = \lim_{x \to 0} \frac{x \cos^2(x)}{x(1 + \frac{\tan(3x)}{x})} ]

Cancel out the common factor of ( x ) in the numerator and denominator: [ \lim_{x \to 0} \frac{\cos^2(x)}{1 + \frac{\tan(3x)}{x}} ]

Substitute ( x = 0 ) into the simplified expression: [ \frac{\cos^2(0)}{1 + \frac{\tan(0)}{0}} = \frac{1}{1 + 0} = 1 ]
Therefore, the limit ( \lim_{x \to 0} \frac{x \cos^2(x)}{x + \tan(3x)} ) without using L'Hôpital's rule is 1.
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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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