Dear student, limt→0bt+1-at+1b-a1tlimit is in form of 1∞ use limx→0 fxgx =elimx→0 gxfx-1=limt→01tbt+1-at+1b-a-1=limt→0bt+1-at+1-b+atb-a00 form so use L'Hospital's Rule=limt→0lnb bt+1-lna at+1b-a = blnb-alna b-a Regards

Q 9 LIMIT

Q 9 LIMIT a and b (b a) numbers then the value Of Lim has the alnb—b/na b/nb—a/na Consider function f : R — I—I, I R. f(x) (A) it is continuous at the (B) it is not derivable at the origin. (C) the range of the function is R. increasing everywhere in its domain. . Then the incorrect statement is I I. Lim x] I then { I } is equal to (wheæ and { } denotes geatest integer and fractional part function.) 2 Which one Of the following limit does not tends to unity? 2 (A) Limit sin (tan t) (B) Limit Sin (cosx) Limit x (D) cosx 13. 14. 15. Limit (A) e -1/12 Limit 3 has the value equal to : (B) e-1/6 . cot-I is equal to (C) e-lß -1 2x+l (D) non existent Let f(x) be a differentiable function which satisfies the equation f(xy) = f(x) + f(y) for all x > O, y > O then f' (x) is equal to x 16. The valueof lim ( x-+0 cos ax 8b2 msec2bx is 2a2

Dear student,

\lim_{t \to 0} {(\frac{b^{t + 1} - a^{t + 1}}{b - a})}^{\frac{1}{t}} l i m i t i s i n f o r m o f 1^{\infty} u s e \lim_{x \to 0} {(f (x))}^{(g (x))} = e^{\lim_{x \to 0} g (x) (f (x) - 1)} = \lim_{t \to 0} \frac{1}{t} (\frac{b^{t + 1} - a^{t + 1}}{b - a} - 1) = \lim_{t \to 0} (\frac{b^{t + 1} - a^{t + 1} - b + a}{t (b - a)}) \frac{0}{0} f o r m s o u s e L' H o s p i t a l' s R u l e = \lim_{t \to 0} (\frac{\ln b (b^{t + 1}) - \ln a (a^{t + 1})}{(b - a)}) = \frac{b \ln b - a \ln a}{(b - a)}

Regards

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