Question

\[\lim_{x \rightarrow 0} ( 1+x+f(x)/x) ) ^{1/x} = e^3\] then find \[\lim_{x \rightarrow 0} \frac{ f(x) }{ x^2}\]

Answer 1

@satellite73 @Mertsj @zepdrix @eashmore @bahrom7893

Answer 2

looks like it would be 2

Answer 3

Yeah ur answer is correct can u Xplain?

Answer 4

@amistre64 @phi @AravindG

Answer 5

Do you know the general expression

lim x-->0 (1+ax)^{1/x} ?

Answer 6

taking log ryt?

Answer 7

Nope. Its ready made formula. It equals e^a

Answer 8

I hope you get the idea from that.

Answer 9

Let \(g(x)=\dfrac{f(x)}{x}\)

then you want

\[\lim_{x\to0}\frac{g(x)}{x}\]

take the log of the original and you have

\[\lim_{x\to 0}\frac{\ln(1+x+g(x))}{x}=3\]

use L'hospitals rule

\[3=\lim_{x\to 0}\frac{\ln(1+x+g(x))}{x}=\lim_{x\to 0}\frac{1+g'(x)}{1+x+g(x)}\]

clearly \(1+x+g(x)\to 1\) otherwise the original limit would not be exp(3)

so we have \(3=\displaystyle\lim_{x\to 0}(1+g'(x))\)

so

\[\lim_{x\to 0}g'(x)=2\]


then \[\lim_{x\to0}\frac{f(x)}{x^2}=\lim_{x\to0}\frac{g(x)}{x}=\lim_{x\to 0}\frac{g'(x)}{1}=2\]

Answer 10

Here is another way to do this. According to
https://en.wikipedia.org/wiki/Exponential_function#Formal_definition
\[e^x= \lim_{n \rightarrow \infty} \left( 1+\frac{ x }{ n } \right)^n\]
replace x with 3 and do a change of variables n =1/x to get
\[ e^3= \lim_{x \rightarrow 0} \left( 1+3x \right)^\frac{1}{x}\]
pattern match this with your problem
\[ \lim_{x \rightarrow 0} \left( 1+x+\frac{f(x)}{x}\right ) ^{1/x} = e^3 \]
and we see that
\[ \lim_{x \rightarrow 0}\left( x+\frac{f(x)}{x}=3x\right )\]
divide by x :
\[ \lim_{x \rightarrow 0}\left( 1+\frac{f(x)}{x^2}=3\right ) \\
\lim_{x \rightarrow 0}\left( \frac{f(x)}{x^2}=2\right ) \\
\lim_{x \rightarrow 0}\frac{f(x)}{x^2}=2
\]