A calculus problem by Simran Bajaj

Calculus Level 3

What is the value of f ( 0 ) f(0) such that the function f : [ 0 , ∞ ) → R f: [0, \infty) \rightarrow \mathbb{R} given by f ( x ) = ( 1 + tan ⁡ 2 x ) 1 / ( 2 x ) f(x)=( 1+ \tan^2 \sqrt x)^{1/(2x)} is continuous everywhere?

0 0 1 1 e − 1 / 2 e^{-1/2} e 1 / 2 e^{1/2}

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Simran Bajaj by Simran Bajaj , 22, India

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2 solutions

Nowras Otmen
Nov 12, 2016

To find the value for which the function is continous at 0, we evaluate the limit lim ⁡ x → 0 + f ( x ) \lim_{x\rightarrow 0^{+}}f(x) .

lim ⁡ x → 0 + f ( x ) = lim ⁡ x → 0 + ( 1 + tan ⁡ 2 x ) 1 / ( 2 x ) \lim_{x\rightarrow 0^{+}}f(x)=\lim_{x\rightarrow 0^{+}}(1+\tan^2 \sqrt{x})^{1/(2x)}

Since sec ⁡ 2 x = 1 + tan ⁡ 2 x \sec^2 x=1+\tan^2 x , we can rewrite the limit as

lim ⁡ x → 0 + ( 1 + tan ⁡ 2 x ) 1 / ( 2 x ) = lim ⁡ x → 0 + ( sec ⁡ 2 x ) 1 / ( 2 x ) \lim_{x\rightarrow 0^{+}}(1+\tan^2 \sqrt{x})^{1/(2x)}=\lim_{x\rightarrow 0^{+}}(\sec^2 \sqrt x)^{1/(2x)}

Now, we consider that ( sec ⁡ 2 x ) 1 / ( 2 x ) = exp ( 2 ln ⁡ sec ⁡ x 2 x ) (\sec^2 \sqrt x)^{1/(2x)}=\text{exp}\left(\cfrac{2\ln \sec \sqrt x}{2x}\right) and once again rewrite the limit

lim ⁡ x → 0 + ( sec ⁡ 2 x ) 1 / ( 2 x ) = lim ⁡ x → 0 + exp ( 2 ln ⁡ sec ⁡ x 2 x ) \lim_{x\rightarrow 0^{+}}(\sec^2 \sqrt x)^{1/(2x)}=\lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{2\ln \sec \sqrt x}{2x}\right)

Now, we use L'Hopital's repeatedly and apply the limit

lim ⁡ x → 0 + exp ( ln ⁡ sec ⁡ x x ) = lim ⁡ x → 0 + exp ( sec ⁡ x tan ⁡ x 2 x sec ⁡ x ) = lim ⁡ x → 0 + exp ( tan ⁡ x 2 x ) \lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\ln \sec \sqrt x}{x}\right)=\lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\sec \sqrt {x} \tan \sqrt {x}}{2\sqrt {x} \sec \sqrt x}\right)=\lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\tan \sqrt x}{2\sqrt x}\right)

lim ⁡ x → 0 + exp ( tan ⁡ x 2 x ) = lim ⁡ x → 0 + exp ( sec ⁡ 2 x 2 ⋅ 2 x 2 x ) = lim ⁡ x → 0 + exp ( sec ⁡ 2 x 2 ) = exp ( 1 / 2 ) = e 1 / 2 = f ( 0 ) . \lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\tan \sqrt x}{2\sqrt x}\right)=\lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\sec^2 \sqrt x}{2\cdot \cfrac{2\sqrt x}{2\sqrt x}}\right)=\lim_{x\rightarrow 0^{+}}\text{exp}\left(\cfrac{\sec^2 \sqrt x}{2}\right)=\text{exp}(1/2)=e^{1/2}=f(0).

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Tapas Mazumdar
Feb 4, 2017

Since we need to find the value of f ( 0 ) f(0) for which the function is continuous everywhere, we should find the limit of f ( x ) f(x) as x x approaches 0 0 from the positive side.

Thus

f ( x ) = ( 1 + tan ⁡ 2 x ) 1 / ( 2 x ) ⟹ ln ⁡ f ( x ) = ln ⁡ ( 1 + tan ⁡ 2 x ) 2 x ⟹ lim ⁡ x → 0 + ln ⁡ f ( x ) = lim ⁡ x → 0 + ln ⁡ ( 1 + tan ⁡ 2 x ) 2 x = lim ⁡ x → 0 + 1 1 + tan ⁡ 2 x ⋅ 2 tan ⁡ x ⋅ sec ⁡ 2 x ⋅ 1 2 x 2 Using L’ Hopital’s Rule for the 0 0 case here = lim ⁡ x → 0 + 1 2 ⋅ tan ⁡ x x = 1 2 As lim ⁡ θ → 0 tan ⁡ θ θ = 1 ⟹ lim ⁡ x → 0 + f ( x ) = e 1 / 2 \begin{aligned} & f(x) &=& {(1+\tan^2 \sqrt{x})}^{1/{(2x)}} \\ \implies & \ln f(x) &=& \dfrac{\ln (1+\tan^2 \sqrt{x})}{2x} \\ \implies & \displaystyle \lim_{x \to 0^{+}} \ln f(x) &=& \lim_{x \to 0^{+}} \dfrac{\ln (1+\tan^2 \sqrt{x})}{2x} \\ & &=& \lim_{x \to 0^{+}} \dfrac{\frac{1}{1+ \tan^2 \sqrt{x}} \cdot 2 \tan \sqrt{x} \cdot \sec^2 \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}}{2} & & \small \color{#3D99F6}{\text{Using L' Hopital's Rule for the } \dfrac 00 \text{ case here}} \\ & &=& \lim_{x \to 0^{+}} \dfrac 12 \cdot \dfrac{\tan \sqrt{x}}{\sqrt{x}} \\ & &=& \dfrac 12 & & \small \color{#3D99F6}{\text{As } \lim_{\theta \to 0} \dfrac{\tan \theta}{\theta} = 1} \\ \implies & \lim_{x \to 0^{+}} f(x) &=& \boxed{e^{1/2}} \end{aligned}

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But isn't 1 ∞ = 1 1^{\infty}=1 as we put zero we get that.

Sahil Silare - 4 years, 4 months ago

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Ahha! A common misconception. Let me show you why 1 ∞ 1^{\infty} is an indeterminate from.

First suppose

1 ∞ = g ( t ) h ( t ) 1^{\infty} = {g(t)}^{h(t)}

where g ( t ) g(t) and h ( t ) h(t) are two different functions which give 1 1 and ∞ \infty for some particular t t , eg.- g ( t ) = 1 + t g(t) = 1+t and h ( t ) = ∣ 1 t ∣ h(t) = \left|\dfrac 1t \right| (here t = 0 t=0 ).

Now, we'll manipulate some things here

g ( t ) h ( t ) = e ln ⁡ ( g ( t ) h ( t ) ) = exp { h ( t ) ln ⁡ ( g ( t ) ) } = exp { ln ⁡ ( g ( t ) ) 1 h ( t ) } \begin{aligned} {g(t)}^{h(t)} &=& e^{\ln \left({g(t)}^{h(t)} \right)} \\ &=& \text{exp} \left\{ h(t) \ln \left( g(t) \right) \right\} \\ &=& \text{exp} \left\{ \dfrac{\ln \left( g(t) \right)}{\frac{1}{h(t)}} \right\} \end{aligned}

Putting back our value of g ( t ) g(t) and h ( t ) h(t) , we get

1 ∞ = exp ( 0 0 ) 1^{\infty} = \text{exp} \left( \dfrac 00 \right)

Thus, we arrive e e raised to the indeterminate form of 0 / 0 0/0 .

However, for lim ⁡ n → ∞ 1 n \displaystyle \lim_{n \to \infty} 1^n the answer is 1 1 .

Tapas Mazumdar - 4 years, 4 months ago

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