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lim_{x rightarrow 1}sqrt{x+3}-2over lim_{x rightarrow 1}sqrt{10-x}-3
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\[\lim_{x \rightarrow 1}\sqrt{x+3}-2\over \lim_{x \rightarrow 1}\sqrt{10-x}-3\]
0 \or 0/0
no thats not the answer
-1 final answer
how do you get that?
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when you initially evaluate a lim and get an indeterminate answer such as 0/0 use L'hopitals rule which says an equivalent limit can be obtained by differentiating the top and bottom then re-evaluating After differentiating: \[\lim_{x \rightarrow 1} \frac{\frac{1}{2\sqrt{x+3}}}{-\frac{1}{2\sqrt{10-x}}} = \lim_{x \rightarrow 1} -\frac{\sqrt{10-x}}{\sqrt{x+3}} = -\frac{3}{2}\]
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