Question

If f(x) = x^3 + 3x^2 + Sin(x) and g(x) = e^x – 1 than find value of \(\lim_{x\rightarrow 0}f(x)^{\frac{1}{g(x)}}\)

(a) e^6e

(b) e^(e/6)

(c) e^6

(d) e^(6/e)

I had been asked this question in class test.

Question is taken from Indeterminate Forms in section Differential Calculus of Engineering Mathematics

Answer 1

Right option is (c) e^6

The explanation is: y=\(lim_{x\rightarrow 0}f(x)^{(\frac{1}{g(x)})} = lim_{x\rightarrow 0}[x^3+3x^2+sin(x)]^{\frac{1}{e^x-1}}\)

Taking ln on both side

ln(y)=\(lim_{x\rightarrow 0}\frac{ln(x^3+3x^2+sin(x))}{e^x-1}\)=ln(0)/0 (Indeterminate Form)

By L’Hospital Rule

\(lim_{x\rightarrow 0}\frac{3x^+6x+cos(x)}{e^x(x^3+3x^2+sin(x))}=1/0\)

Again using L’Hospital rule

ln(y)=\(\lim_{x\rightarrow 0}\frac{6x+6+sin(x)}{e^x(x^3+3x^2+sin(x))+e^x(3x^2+6x+cos(x))}=6\)

y=e^6