NDA 2014
Exam NDA
Year 2014
Subject Mathematics
Chapter Limits NDA 2014 · Mathematics · Limits
What is _ x arrow 0 ((1+x)^n-1)/(x) equal to?
What is
lim x → 0 ( 1 + x ) n − 1 x \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} lim x → 0 x ( 1 + x ) n − 1 equal to?
Options A 0 B 1 C n n n D n − 1 n-1 n − 1
Step-by-step solution
lim x → 0 ( 1 + x ) n − 1 x ( 1 + x ) 1 − 1 x = 1 + x − 1 x = x x = 1 ( 1 + x ) 2 − 1 x = 1 + 2 x + x 2 − 1 x = x ( 2 + x ) x \begin{aligned} & \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} \\ & \frac{(1+x)^1-1}{x}=\frac{1+x-1}{x}=\frac{x}{x}=1 \\ & \frac{(1+x)^2-1}{x}=\frac{1+2 x+x^2-1}{x}=\frac{x(2+x)}{x}\end{aligned} x → 0 lim x ( 1 + x ) n − 1 x ( 1 + x ) 1 − 1 = x 1 + x − 1 = x x = 1 x ( 1 + x ) 2 − 1 = x 1 + 2 x + x 2 − 1 = x x ( 2 + x ) ( 1 + x ) 3 − 1 x = 1 + 3 x + 3 x 2 + x 3 − 1 x = 3 x + 3 x 2 + x 3 x = x ( 3 + 3 x + x 2 ) x ∴ lim x → 0 ( 1 + x ) n − 1 x = 1 + n ⋅ x + … − 1 x = n x + highest power of x x = x ( n + highest power of x ) x = n + highest power of x = n \begin{aligned} \frac{(1+x)^3-1}{x} & =\frac{1+3 x+3 x^2+x^3-1}{x} \\ & =\frac{3 x+3 x^2+x^3}{x}=\frac{x\left(3+3 x+x^2\right)}{x} \\ \therefore \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} & =\frac{1+n \cdot x+\ldots-1}{x} \\ & =\frac{n x+\text{ highest power\text{ of } } x}{x} \\ & =\frac{x(n+\text{ highest power of } x)}{x} \\ & =n+\text{ highest power of } x \\ & =n\end{aligned} x ( 1 + x ) 3 − 1 ∴ x → 0 lim x ( 1 + x ) n − 1 = x 1 + 3 x + 3 x 2 + x 3 − 1 = x 3 x + 3 x 2 + x 3 = x x ( 3 + 3 x + x 2 ) = x 1 + n ⋅ x + … − 1 = x n x + highest power of x = x x ( n + highest power of x ) = n + highest power of x = n $
Practice Limits on Quantrex Academy →
← Previous question What is _ x arrow 0 (5^x-1)/(x) equal to?Next question → Paragraph : Consider the function f(x)=(1- x)/((π-2 x)^2) , where x ≠ (π)/(2) and…
Quantrex AcademyJEE · NEET · NDA PYQs with solutions
Open app