Dear Student,We have to find,limx→1(x-1x3-1)As it is of 00 form Apply L'Hospital's Rulelimx→1(12x13x23)limx→1(3x162)=32limx→0 (1+x-1x)First rationalize the numerator1+x-1x(1+x+11+x+1)=1+x-1x(1+x+1)=11+x+1limx→011+x+1=1Regards

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D e a r S t u d e n t, W e h a v e t o f i n d, \lim_{x \to 1} (\frac{\sqrt{x} - 1}{\sqrt[3]{x} - 1}) A s i t i s o f \frac{0}{0} f o r m . A p p l y L' H o s p i t a l' s R u l e \lim_{x \to 1} (\frac{\frac{1}{2 \sqrt{x}}}{\frac{1}{3 x^{\frac{2}{3}}}}) \lim_{x \to 1} (\frac{3 x^{\frac{1}{6}}}{2}) = \frac{3}{2} \lim_{x \to 0} (\frac{\sqrt{1 + x} - 1}{x}) F i r s t r a t i o n a l i z e t h e n u m e r a t o r \frac{\sqrt{1 + x} - 1}{x} (\frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}) = \frac{1 + x - 1}{x (\sqrt{1 + x} + 1)} = \frac{1}{\sqrt{1 + x} + 1} \lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} = 1 R e g a r d s

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