andrey21
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Use L'Hospital's rule to find the limit of the following:
(SQRT (3+x) -2)/(x-1)
As x tends to 1
Here's my attempt
Let f(x) = SQRT(3+x) - 2 g(x) = x-1
f'(x) = 1/2 (SQRT(3+x)) g'(x) = 1
Therefore f'(x)/g'(x) is:
1/2(SQRT(3+x)/1
=1/2(SQRT(3+1)
= 1/2(SQRT(4))
=1/4
(SQRT (3+x) -2)/(x-1)
As x tends to 1
Here's my attempt
Let f(x) = SQRT(3+x) - 2 g(x) = x-1
f'(x) = 1/2 (SQRT(3+x)) g'(x) = 1
Therefore f'(x)/g'(x) is:
1/2(SQRT(3+x)/1
=1/2(SQRT(3+1)
= 1/2(SQRT(4))
=1/4