# How do you find the limit of #x Tan(9/x)# as x approaches infinity using l'hospital's rule?

See below

In order to use l'Hospital's rule we rewrite the expression:

Differentiating the numerator and denominator gets us:

We do not need l'Hospital

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To find the limit of x Tan(9/x) as x approaches infinity using L'Hospital's rule, we can apply the rule to the expression.

First, we differentiate both the numerator and denominator with respect to x. The derivative of x is 1, and the derivative of Tan(9/x) can be found using the chain rule.

Applying the chain rule, we have: d/dx(Tan(9/x)) = sec^2(9/x) * (-9/x^2)

Now, we can rewrite the expression as: lim(x->∞) [1 / (sec^2(9/x) * (-9/x^2))]

Next, we simplify the expression by taking the reciprocal of the denominator and multiplying it by the numerator: lim(x->∞) [(-x^2) / (9 * sec^2(9/x))]

As x approaches infinity, the term 9/x approaches 0. Therefore, we can rewrite the expression as: lim(x->∞) [(-x^2) / (9 * sec^2(0))]

Since sec^2(0) is equal to 1, the expression simplifies further to: lim(x->∞) [(-x^2) / 9]

As x approaches infinity, the numerator (-x^2) also approaches infinity. Therefore, the limit of x Tan(9/x) as x approaches infinity using L'Hospital's rule is negative 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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