# How do you find #lim (xlnx)/(x^2-1)# as #x->1# using l'Hospital's Rule?

Let's put:

As:

and

we can use l'Hopital's rule stating that in such case:

if such limits exists.

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To find ( \lim_{x \to 1} \frac{x \ln x}{x^2 - 1} ) using L'Hôpital's Rule:

- Evaluate the limit directly at ( x = 1 ) to check if it is an indeterminate form.

[ \lim_{x \to 1} \frac{x \ln x}{x^2 - 1} = \frac{1 \cdot \ln 1}{1^2 - 1} = \frac{0}{0} ]

- Apply L'Hôpital's Rule by taking the derivative of the numerator and the derivative of the denominator separately.

[ \lim_{x \to 1} \frac{x \ln x}{x^2 - 1} = \lim_{x \to 1} \frac{\frac{d}{dx}(x \ln x)}{\frac{d}{dx}(x^2 - 1)} ]

- Find the derivatives:

[ \lim_{x \to 1} \frac{\frac{d}{dx}(x \ln x)}{\frac{d}{dx}(x^2 - 1)} = \lim_{x \to 1} \frac{(\ln x + 1)}{(2x)} ]

- Evaluate the limit again:

[ \lim_{x \to 1} \frac{(\ln x + 1)}{(2x)} = \frac{(\ln 1 + 1)}{(2 \cdot 1)} = \frac{1}{2} ]

Therefore, ( \lim_{x \to 1} \frac{x \ln x}{x^2 - 1} = \frac{1}{2} ).

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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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