How do you find #lim ((1-x)^(1/4)-1)/x# as #x->0# using l'Hospital's Rule?
Take the derivative of the top and bottom. Don't forget chain rule:
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Alt approach:
By Binomial Series:
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To find ( \lim_{{x \to 0}} \frac{{(1-x)^{\frac{1}{4}} - 1}}{x} ) using L'Hôpital's Rule, you first notice that this expression is of the form ( \frac{0}{0} ) as ( x \to 0 ). So, you can apply L'Hôpital's Rule.
Differentiate the numerator and the denominator with respect to ( x ), then evaluate the limit again.
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Differentiate the numerator: [ \frac{d}{dx}((1-x)^{\frac{1}{4}} - 1) = \frac{1}{4}(1-x)^{-\frac{3}{4}} \cdot (-1) \cdot (-1) = \frac{1}{4(1-x)^{\frac{3}{4}}} ]
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Differentiate the denominator: [ \frac{d}{dx}(x) = 1 ]
Now, take the limit again as ( x \to 0 ) with the new expressions obtained after differentiation.
[ \lim_{{x \to 0}} \frac{\frac{1}{4(1-x)^{\frac{3}{4}}}}{1} = \frac{1}{4} \lim_{{x \to 0}} \frac{1}{(1-x)^{\frac{3}{4}}} ]
Now, plug ( x = 0 ) into the expression:
[ \frac{1}{4} \cdot \frac{1}{(1-0)^{\frac{3}{4}}} = \frac{1}{4} ]
So, ( \lim_{{x \to 0}} \frac{{(1-x)^{\frac{1}{4}} - 1}}{x} = \frac{1}{4} ).
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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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