Let f be an invertible function defined on R such that its inverse function f -1( x )= x 3+4 x 2+6 x +1, then the value of undersetx arrow 1 textLim (f(x)+x)(1/x2+5 x-6) equals

Question

Let f be an invertible function defined on R such that its inverse function f -1( x )= x 3+4 x 2+6 x +1, then the value of undersetx arrow 1 textLim (f(x)+x)(1/x2+5 x-6) equals (A) (1/ e 1 / 6) (B) e 1 / 6 (C) e 6 (D) e -6

Tardigrade Admin · Accepted Answer

f -1(0)=1 ⇒ f (1)=0 ∴ L= undersetx arrow 1 textLim (f(x)+x)(1/x2+5 x-6)=e operatornameLimx arrow 1 (f(x)+x-1/x2+5 x 6)=e(f prime(1)+1/7) Now, f prime(1)=(1/(f-1) prime(0))=(1/6) ⇒ L = e 1 / 6

Q. Let $f$ be an invertible function defined on $R$ such that its inverse function $f^{- 1} (x) = x^{3} + 4 x^{2} + 6 x + 1$ , then the value of $x \to 1 Lim (f (x) + x)^{\frac{1}{x ^{2} + 5 x - 6}}$ equals

$\frac{1}{e ^{1/6}}$

$e^{1/6}$

$e^{6}$

$e^{- 6}$

$f^{- 1} (0) = 1 \Rightarrow f (1) = 0$
$∴ L = x \to 1 Lim (f (x) + x)^{\frac{1}{x ^{2} + 5 x - 6}} = e^{Lim_{x \to 1} \frac{f ( x ) + x - 1}{x ^{2} + 5 x 6}} = e^{\frac{f ^{'} ( 1 ) + 1}{7}}$
Now, $f^{'} (1) = \frac{1}{( f ^{- 1} ) ^{'} ( 0 )} = \frac{1}{6}$
$\Rightarrow L = e^{1/6}$

Q. Let f be an invertible function defined on R such that its inverse function f−1(x)=x3+4x2+6x+1, then the value of x→1Lim​(f(x)+x)x2+5x−61​ equals

e1/61​

e1/6

e6

e−6