Verified Solution • 2025–26 Session moderate

QJEEM17Ma7c17: JEE Main 2017 Mathematics - Class 12 JEEM17Ma7c17 Solution

lim⁡x→3\mathop {\lim }\limits_{x \to 3} 3x−32x−4−2{{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }} is equal to :

Multiple Choice Options

  • A 3\sqrt 3
  • B 12{1 \over {\sqrt 2 }}
  • C 32{{\sqrt 3 } \over 2}
  • D 122{1 \over {2\sqrt 2 }}

Step-by-Step Verified Solution

Given,

       lim⁡x→3\mathop {\lim }\limits_{x \to 3} 3x−32x−4−2{{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}

Here if you put x = 3 in 3x−32x−4−2{{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}

you will get 00{0 \over 0} form.

So, we can apply L' Hospital rule

∴   \therefore\,\,\, lim⁡x→33x−32x−4−2\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}

= lim⁡x→3\mathop {\lim }\limits_{x \to 3} 3.12x222x−4{{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}} (applying L' Hospital rule)

= 3.12316−4{{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}

= 12{1 \over 2} ×\times 2\sqrt 2

= 12{1 \over {\sqrt 2 }}

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