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\(\rm \displaystyle\lim_{x\rightarrow 2}\frac{x^2-3x+2}{\sqrt{x}-\sqrt{2}}\) is equal to
5. 1

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Correct Answer - Option 2 : \(2 \sqrt{2} \)


A fraction whose numerator and denominator both tend to zero as x → a is an example of an indeterminate form written as 0/0. it has no definite values. other indeterminate forms are: ∞/∞, ∞ - ∞, 0 x ∞, 1, 00, ∞0. indeterminate form are not any definite number and hence are not acceptable as limits. to find limits in such cases, we use the L'hospital's rule, rationalization method, dividing the  numerator and denominator by the higher power or factorization method. 


\(\rm \displaystyle\lim_{x\rightarrow 2}\frac{x^2-3x+2}{\sqrt{x}-\sqrt{2}}\) This is 0/0 form.

Here we use  factorization method

\(\rm \displaystyle\lim_{x\rightarrow 2}\frac{(x-2)(x-1)}{\sqrt{x}-\sqrt{2}}\)

\(\rm \displaystyle\lim_{x\rightarrow 2}\frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})(x-1)}{\sqrt{x}-\sqrt{2}}\)

\(\rm \displaystyle\lim_{x\rightarrow 2}{(\sqrt{x}+\sqrt{2})(x-1)} = \rm {(\sqrt{2}+\sqrt{2})(2-1)} = 2\sqrt{2}\)

Hence, option 2 is the correct answer.

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