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calculate \mathrm{\lim _{x \rightarrow 3} \frac{\sqrt{19-x}-2 \sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}}

Option: 1

\frac{9}{52}


Option: 2

\frac{1}{52}


Option: 3

\frac{1}{2}


Option: 4

0


Answers (1)

best_answer

\mathrm{\begin{aligned} \frac{\sqrt{19-x}-2 \sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1} & =\frac{\left((19-x)^2\right)^{\frac{1}{4}}-\left(2^4(13+x)\right)^{\frac{1}{4}}}{(11-x)^{\frac{1}{3}}-\left((x-1)^3\right)^{\frac{1}{3}}} \\ & =\frac{a^{\frac{1}{4}}-b^{\frac{1}{4}}}{c^{\frac{1}{3}}-d^{\frac{1}{3}}} \\ & =\frac{a^{\frac{1}{4}}-b^{\frac{1}{4}}}{c^{\frac{1}{3}}-d^{\frac{1}{3}}} \times \frac{a^{\frac{2}{4}}+a^{\frac{1}{3}} b^{\frac{1}{4}}+a^{\frac{1}{4}} b^{\frac{1}{4}}+b^{\frac{3}{4}}}{a^{\frac{2}{4}}+a^{\frac{1}{2}} b^{\frac{1}{4}}+a^{\frac{1}{4}} b^{\frac{1}{2}}+b^{\frac{3}{4}}} \times \frac{c^{\frac{2}{3}}+c^{\frac{1}{3}} d^{\frac{1}{3}}+d^{\frac{2}{3}}}{c^{\frac{2}{3}}+c^{\frac{1}{3}} d^{\frac{1}{2}}+d^{\frac{2}{3}}} \end{aligned}}

                                             \mathrm{\begin{aligned} & =\frac{\left(a^{\frac{1}{4}}\right)^4-\left(b^{\frac{1}{4}}\right)^4}{\left(c^{\frac{1}{3}}\right)^3-\left(d^{\frac{1}{3}}\right)^3} \times \frac{c^{\frac{2}{3}}+c^{\frac{1}{3}} d^{\frac{1}{4}}+d^{\frac{2}{3}}}{a^{\frac{3}{4}}+a^{\frac{1}{2}} b^{\frac{1}{4}}+a^{\frac{1}{4}} b^{\frac{1}{4}}+b^{\frac{3}{4}}} \\ & =\frac{a-b}{c-d} \times \frac{c^{\frac{2}{3}}+c^{\frac{1}{3}} d^{\frac{1}{3}}+d^{\frac{2}{3}}}{a^{\frac{3}{4}}+a^{\frac{1}{2}} b^{\frac{1}{4}}+a^{\frac{1}{4}} b^{\frac{1}{2}}+b^{\frac{3}{4}}} \end{aligned}}

(where I hope the replacements of a,b,c,d are obvious )

Upon simplification (with \mathrm{(x \neq 3)} ,we have

\mathrm{\frac{a-b}{c-d}=\frac{153-54 x+x^2}{12-4 x+3 x^2-x^3}=\frac{(51-x)(3-x)}{(3-x)\left(4+x^2\right)}=\frac{51-x}{4+x^2}}

and the remaining fraction of rational powers is continuous at x= 3, Then the limit is 

\mathrm{\lim _{x \rightarrow 3} \frac{\sqrt{19-x}-2 \sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13} \times \frac{12}{256}=\frac{9}{52}}

Posted by

Riya

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