How do you find #lim (x+5)(1/(2x)+1/(x+2))# as #x->0^+# using l'Hospital's Rule or otherwise?

To find ( \lim_{x \to 0^+} (x+5)\left(\frac{1}{2x}+\frac{1}{x+2}\right) ), we can use l'Hôpital's Rule or simplify the expression.

Using l'Hôpital's Rule:

[ \lim_{x \to 0^+} (x+5)\left(\frac{1}{2x}+\frac{1}{x+2}\right) = \lim_{x \to 0^+} \frac{(x+5)}{2x} + \lim_{x \to 0^+} \frac{(x+5)}{x+2} ]

[ = \lim_{x \to 0^+} \frac{1}{2} + \lim_{x \to 0^+} \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1 ]

Alternatively, simplifying the expression:

[ (x+5)\left(\frac{1}{2x}+\frac{1}{x+2}\right) = (x+5)\left(\frac{x+2+2x}{2x(x+2)}\right) ]

[ = (x+5)\left(\frac{3x+2}{2x(x+2)}\right) = \frac{(3x+2)(x+5)}{2x(x+2)} ]

[ = \frac{3x^2+17x+10}{2x(x+2)} ]

Taking the limit as (x) approaches (0^+):

[ \lim_{x \to 0^+} \frac{3x^2+17x+10}{2x(x+2)} = \frac{10}{0^+} = +\infty ]

However, the first method using l'Hôpital's Rule is simpler and more direct for this problem. Therefore, ( \lim_{x \to 0^+} (x+5)\left(\frac{1}{2x}+\frac{1}{x+2}\right) = 1 ).