How do you find #lim (3x^2+x+2)/(x-4)# as #x->0# using l'Hospital's Rule or otherwise?
The initial form is not indeterminate. We cannot use l'Hospiital for this.
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To find ( \lim_{x \to 0} \frac{3x^2 + x + 2}{x - 4} ), we first try to directly substitute ( x = 0 ) into the expression. However, this results in an indeterminate form of (\frac{2}{-4}). Therefore, we can apply L'Hôpital's Rule by taking the derivatives of the numerator and denominator with respect to ( x ) separately until we get a determinate form.
[ f(x) = 3x^2 + x + 2 ] [ g(x) = x - 4 ]
Taking the derivatives: [ f'(x) = 6x + 1 ] [ g'(x) = 1 ]
Now, applying L'Hôpital's Rule: [ \lim_{x \to 0} \frac{3x^2 + x + 2}{x - 4} = \lim_{x \to 0} \frac{6x + 1}{1} = \frac{1}{1} = 1 ]
So, ( \lim_{x \to 0} \frac{3x^2 + x + 2}{x - 4} = 1 ).
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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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