l i m x → 0 ( 1 + 2 x 1 + 3 x ) 1 x 2 . e 1 x = Options: A . &nbsp e 5 2 B . &nbsp e 2 C . &nbsp 2 D . &nbsp 1

Answer: Option A : A l i m x → 0 e 1 x 2 ( l o g ( 1 + 2 x ) − l o g ( 1 + 3 x ) ) + 1 x e l i m x → 0 ( l o g ( 1 + 2 x ) − l o g ( 1 + 3 x ) ) + x x 2 Using the expansion we get - = e 5 2

limx→0(1+2x1+3x)1x2.e1x=

Discussion Forum : Limits Continuity And Differentiability

Question -

$l i m x \to 0 {(\frac{1 + 2 x}{1 + 3 x})}^{\frac{1}{x^{2}}} . e^{\frac{1}{x}} =$

Options:

A .

e^{\frac{5}{2}}

B .

e^{2}

C . 2

D . 1

Answer: Option A
:
A

l i m x \to 0 e^{\frac{1}{x^{2}} (l o g (1 + 2 x) - l o g (1 + 3 x)) + \frac{1}{x}} e^{l i m x \to 0 \frac{(l o g (1 + 2 x) - l o g (1 + 3 x)) + x}{x^{2}}}

Using the expansion we get -

= e^{\frac{5}{2}}

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Discussion Forum : Limits Continuity And Differentiability

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