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प्रश्न
\[\lim_{x \to a} \frac{x - a}{\sqrt{x} - \sqrt{a}}\]
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उत्तर
\[\lim_{x \to a} \left[ \frac{x - a}{\sqrt{x} - \sqrt{a}} \right]\]
= \[\lim_{x \to a} \left[ \frac{\left( \sqrt{x} \right)^2 - a^2}{\sqrt{x} - \sqrt{a}} \right]\]
= \[\lim_{x \to a} \left[ \frac{\left( \sqrt{x} - \sqrt{a} \right) \left( \sqrt{x} + \sqrt{a} \right)}{\left( \sqrt{x} - \sqrt{a} \right)} \right]\]
= \[\sqrt{a} + \sqrt{a}\]
= \[2\sqrt{a}\]
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संबंधित प्रश्न
Let a1, a2,..., an be fixed real numbers and define a function f ( x) = ( x − a1 ) ( x − a2 )...( x − an ).
What is `lim_(x -> a_1) f(x)` ? For some a ≠ a1, a2, ..., an, compute `lim_(x -> a) f(x)`
If f(x) = `{(|x| + 1,x < 0), (0, x = 0),(|x| -1, x > 0):}`
For what value (s) of a does `lim_(x -> a)` f(x) exists?
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