How do you find #lim (x+5)(1/(2x)+1/(x+2))# as #x->1# using l'Hospital's Rule or otherwise?
Substitution.
Nothing becomes undefined or indeterminate, so use substitution.
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To find ( \lim_{x \to 1} (x + 5)\left(\frac{1}{2x} + \frac{1}{x + 2}\right) ), you can simplify the expression first, then apply direct substitution.
Simplifying the expression, we get: [ (x + 5)\left(\frac{1}{2x} + \frac{1}{x + 2}\right) = \frac{x + 5}{2x} + \frac{x + 5}{x + 2} ]
Now, substitute ( x = 1 ) into the expression: [ \lim_{x \to 1} \left(\frac{x + 5}{2x} + \frac{x + 5}{x + 2}\right) = \frac{1 + 5}{2 \cdot 1} + \frac{1 + 5}{1 + 2} = \frac{6}{2} + \frac{6}{3} = 3 + 2 = 5 ]
Therefore, ( \lim_{x \to 1} (x + 5)\left(\frac{1}{2x} + \frac{1}{x + 2}\right) = 5 ).
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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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