ISRO2016-3

Question

$\displaystyle{}\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$ is given by

• A. $0$
• B. $-1$
• C. $1$
• D. $\frac{1}{2}$

Comments

how r u telling no?

here 0 and 1 both could be the answer

Applying L hospital rule

---

@srestha how?

---

yes rt.

Answer 1

$\lim_{x->0}\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$  [putting x=0, it comes 0/0 form , So, apply L hospital rule(differentiate upper limit and lower limit differently)]

$\lim _{x \rightarrow 0}\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}=\frac{1}{2}+\frac{1}{2}=1$

Comments

You put the incorrect "+" sign between the two differentiated values. It should be $\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}} = \frac{1}{2} - \frac{1}{2} = 0$

---

This is just to share my observation on how sign part changes to '+'

---

ans is 1

after taking derivative of sqrt(1-x)=1/2*sqrt(1-x)*(-1)

so sign changes to +

---

Yeah you are right. I forgot to differentiate the (1+x) and (1-x).

---

this problem can be solved with factorization, no need to apply differentiation here. and it's simple, don't make it complicate.