Question

If f(x) = $1/\sqrt{2}$is continuous at

x = 0, then -

• A. a = $\lim_{x \rightarrow a}\frac{x^{3} - a^{3}}{x^{2} - a^{2}} =$, b = $\frac{3a}{2}$
• B. a = $\lim_{h \rightarrow 0}\frac{\sqrt{x + h} - \sqrt{x}}{h} =$, b = $1/2\sqrt{x}$
• C. a = $1/2\sqrt{h}$ , b = $\lim_{\alpha \rightarrow \beta}\frac{\sin^{2}\alpha - \sin^{2}\beta}{\alpha^{2} - \beta^{2}}$
• D. a = $\frac{\sin\beta}{\beta}$, b = $\frac{\sin 2\beta}{2\beta}$

Answer 1

Answer: (B)

a = $\lim_{h \rightarrow 0}\frac{\sqrt{x + h} - \sqrt{x}}{h} =$, b = $1/2\sqrt{x}$

Answer 2

If continuous at x = 0 then

b = $\lim _ { x \rightarrow 0 }$ ,

b = ea & b = $\lim _ { x \rightarrow 0 }$ $\mathrm { e } ^ { \frac { \tan 2 \mathrm { x } } { \tan 3 \mathrm { x } } }$

⇒ b = e^2/3

so e^a = e^2/3 = b ⇒ a = $\frac { 2 } { 3 }$ & b = e^2/3