Category Archives: Calculus

Solved Problems in Calculus

Problem 2-5: Evaluating Indeterminate Form


Calculate  \displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}.

Solution
 \frac{\sqrt{x+2}-2}{x-2}=\frac{0}{0}
 \displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}= \displaystyle\lim_{x\to 2}\frac{\sqrt{x+2}-2}{x-2}\times \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2}

 =\displaystyle\lim_{x\to 2}\frac{x+2-4}{(x-2) \times (\sqrt{x+2}+2)}
 =\displaystyle\lim_{x\to 2}\frac{x-2}{(x-2) \times (\sqrt{x+2}+2)}

 =\displaystyle\lim_{x\to 2}\frac{1}{(\sqrt{x+2}+2)}=\frac{1}{4}

Problem 2-5

Problem 2-2: Evaluating Derivative of Functions and the Tangent Lines


Find the derivative of  f(x) and the equation of the tangent line at  x_0=-1.

a)  f(x)=x^2
b)  f(x)=x^3+x+1
c)  f(x)=\frac{1}{x}

Solution
The equation of the tangent line at  {x}_{0} is  y = f'(x_0) (x-x_0) + f(x_0) .
a)  f(x)=x^2, f'(x)=2x
 y = f'(x_0) (x-x_0) + f(x_0) = 2x_0 (x-x_0)+x_0^2 = 2x_0 x-x_0^2
 y = -2x-1 .


b)  f(x)=x^3+x+1

 f'(x)=3x^2+1
 y=f'(x_0) (x-x_0) + f(x_0) =(3x_0^2+1) (x-x_0)+x_0^3+x_0+1=(3x_0^2+1)x-2x_0^3+1
 y=4x+3 .

c)  f(x)=\frac{1}{x}=x^{-1}
 f'(x)=-x^{-2}  f'(x)=-\frac{1}{x^2}
 y=f'(x_0) (x-x_0) + f(x_0) =-\frac{1}{x_0^2}(x-x_0)+\frac{1}{x_0}=-\frac{1}{x_0^2} x +\frac{2}{x_0}
 y = -x-2