ISRO 2013- Calculus [Mech]

Question

Lim x $\rightarrow$0  $\frac{x^{2}+ x - Sin x}{x^{2}}$
(a) 0
(b) ∞

(c) 1

(d) None of these

Answer 1

$\frac{0}{0}$ indeterminate form. Apply repeated L'Hospital rule,

$\lim_{x\rightarrow 0}\frac{x^2 + x - sinx}{x^2} = \frac{2x + 1 -cosx}{2x} = \frac{2 + sinx}{2} = 1$

Answer 2

Option C is correct answer.

Lim(x->0) on  ( x^2 + x- Sin(x) ) / x^2

Solve the function first and then apply limit on every part of the function.

  =( x^2 / x^2 ) + ( x / x^2 ) - ( Sin(x) / x^2 )

= 1 + 1/x - ( Sin(x) / x^2 ) ..........eq1

Apply limit on eq1

 Lim(x->0) on 1 + Lim(x->0) on 1/x - Lim(x->0) on ( Sin(x) / x^2 )

=1 +  Lim(x->0) on 1/x - Lim(x->0) on ( Sin(x) / x^2 )

(as we know  Lim(x->0) on Sin(x) / x =1)

=1 + Lim(x->0) on 1/x - Lim(x->0) on 1/x

we get answer = 1