Solve lim x-1 abx-axb/x2-1

limx→1abx-axbx2-1If you put x = 1, you will get 00 form, so Use L-hospital rule Hence differentiating numerator and denominator individually untill this form is removedddx(abx-axb)ddx(x2-1)=abxlogb - baxloga2x, here 00 form is removed, so put x =1, get the limit Hence abxlogb - baxloga2x at x= 1 =ablogb-abloga2Hence limx→1abx-axbx2-1 =ablogb-abloga2 (ans)

Solve lim x-1 ab^x-a^xb/x²-1

\lim_{x \to 1} \frac{a b^{x} - a^{x} b}{x^{2} - 1} I f y o u p u t x = 1, y o u w i l l g e t \frac{0}{0} f o r m, s o U s e L - h o s p i t a l r u l e . H e n c e d i f f e r e n t i a t i n g n u m e r a t o r a n d d e n o m i n a t o r i n d i v i d u a l l y u n t i l l t h i s f o r m i s r e m o v e d \frac{\frac{d}{d x} (a b^{x} - a^{x} b)}{\frac{d}{d x} (x^{2} - 1)} = \frac{a b^{x} \log b - b a^{x} \log a}{2 x}, h e r e \frac{0}{0} f o r m i s r e m o v e d, s o p u t x = 1, g e t t h e l i m i t . H e n c e \frac{a b^{x} \log b - b a^{x} \log a}{2 x} a t x = 1 = \frac{a b \log b - a b \log a}{2} H e n c e \lim_{x \to 1} \frac{a b^{x} - a^{x} b}{x^{2} - 1} = \frac{a b \log b - a b \log a}{2} (a n s)

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