\lim_{x\to2a} \sqrt{x}-\sqrt{2a}+\sqrt{x-2a}/\sqrt{x^{2}-4a^{2}}}=

AP ICET 2017 Shift 2 Paper

Question : 134 of 200

Marks: +1, -0

\lim_{x→2a}\;\;{√{x}-√{2a}+√{x-2a}}/{√{x^{2}-4a^{2}}}=

Solution:

\lim↙{x→2a}\;\;{√{x}-√{2a}+√{x-2a}}/{√{x^{2}-4a^{2}}}

For x = 2a, it is of

0/0

form. Applying L’ Hospital’s rule,

=\lim↙{x→2a}{{1}/{2√{x}}+{1}/{2√{x-2a}}}/{{2x}/{2√{x^{2}-4a^{2}}}}

=\lim↙{x→2a}{√{x-2a}+√{x}}/{2√{x}√{x-2a}}×{√{x^{2}-4a}}/{x}

=\lim↙{x→2a}\;\;{(√{x-2a}+√{x})(√{x+2a})}/{2x√{x}}

={(√{2a})(√{4a})}/{2×2a√{2a}}={2√{a}}/{2×2a}={1}/{2√{a}}

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