Lim x - 0log e1- x - x -A- 1-eB- eC- 1D- - -1

The correct option is C 1 We know, lim_x→ 0log_e(1+x)/x at x=0 the value of given expression takes0/0 form. Substituting the expression of Log_e(1+x)=x x^2/2+x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Single Point Continuity

lim x → 0log ...

Question

$lim x \to 0 \frac{l o g_{e} (1 + x)}{x} =$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

-1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
1

We know, $lim x \to 0 \frac{l o g_{e} (1 + x)}{x} a t x = 0$

the value of given expression takes $\frac{0}{0}$ form.

Substituting the expression of

$L o g_{e} (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} . . . . . .$ in the given expression

$lim x \to 0 \frac{x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - . . . . .}{x}$

$lim x \to 0 (1 - \frac{x}{2} + \frac{x^{2}}{3} - . . . . .)$

Putting x=0

=1-0=1

Option C is correct

Suggest Corrections

Similar questions

lim x \to 0 \frac{l o g_{e} (1 + x)}{3^{x} - 1} =

lim x \to 0 \frac{l o g_{e} (1 + x)}{3^{x} - 1} =

____.

lim x \to 0 \frac{{log}_{e} (1 + x)}{3^{x} - 1} =

lim x \to 0 {\frac{{log}_{e} (1 + x)}{x^{2}} + \frac{x - 1}{x}}

is equal to

Q. If

f (x)

is defined on

(0, 1)

, then the domain of

g (x) = f (e^{x}) + f ({log}_{e} | x |)

Join BYJU'S Learning Program

limx→ 0loge(1+x)x=

The correct option is C 1 We know, limx→ 0loge(1+x)x at x=0 the value of given expression takes00 form. Substituting the expression of Loge(1+x)=x−x22+x33...... in the given expression limx→ 0x−x22+x33−.....x limx→ 0(1−x2+x23−.....) Putting x=0 =1-0=1 Option C is correct

$lim x \to 0 \frac{l o g_{e} (1 + x)}{x} =$