# How do you find the limit of #x^sqrtx# as x approaches 0 using l'hospital's rule?

1

start by using log properties to get that exponent down

so we use L'Hopital on that :

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To find the limit of x^sqrt(x) as x approaches 0 using L'Hospital's rule, we can rewrite the expression as e^(sqrt(x) * ln(x)). Taking the natural logarithm of both sides, we get ln(x^sqrt(x)) = sqrt(x) * ln(x). Now, we can apply L'Hospital's rule by differentiating the numerator and denominator separately. Differentiating sqrt(x) gives us 1/(2*sqrt(x)), and differentiating ln(x) gives us 1/x. Taking the limit as x approaches 0, we have (1/(2*sqrt(x))) / (1/x). Simplifying this expression, we get (x/(2*sqrt(x))) * (x/1). Canceling out x, we are left with 1/(2*sqrt(x)). Taking the limit as x approaches 0, we find that the limit of x^sqrt(x) as x approaches 0 is 0.

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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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- What is the limit of #(sqrt x) / (x + 4)# as x approaches infinity?
- How do you evaluate the limit #(2sinx-sin2x)/(xcosx)# as x approaches #0#?

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