The correct option is B is e^2For h gt;0 Limit can be re written as h → 0lim(e^h+h)^1/h Let L=(e^h+h)^1/h Taking ln both sides ⇒ln L=1hln(e^h+h) Taking limi ...

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Standard XII

Mathematics

Derivative from First Principle

lim x → 0+ex+...

Question

$lim x \to 0 + (e^{x} + x)^{1 / x}$

Does not exist finitely

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e^{2}

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Solution

The correct option is C is $e^{2}$
For $h > 0$
Limit can be re-written as
$lim h \to 0 (e^{h} + h)^{1 / h}$
Let $L = (e^{h} + h)^{1 / h}$
Taking $ln$ both sides
$\Rightarrow ln L = \frac{1}{h} ln (e^{h} + h)$

Taking limit both sides
$\Rightarrow lim h \to 0 ln L = lim h \to 0 \frac{1}{h} ln (e^{h} + h)$

Using L'Hospital
$\Rightarrow lim h \to 0 ln L = lim h \to 0 \frac{e^{h} + 1}{(e^{h} + h)}$
$\Rightarrow lim h \to 0 ln L = 2$
$\Rightarrow lim h \to 0 L = e^{2}$

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limx→0+(ex+x)1/x

The correct option is C is e2For h>0 Limit can be re-written as limh→0(eh+h)1/h Let L=(eh+h)1/h Taking ln both sides ⇒lnL=1hln(eh+h) Taking limit both sides ⇒limh→0lnL=limh→01hln(eh+h) Using L'Hospital ⇒limh→0lnL=limh→0eh+1(eh+h) ⇒limh→0lnL=2 ⇒limh→0L=e2

$lim x \to 0 + (e^{x} + x)^{1 / x}$