The value of lim h - 0ln 1-2 h-2 ln 1-h-h2 isA- 1B- 0C- -1D- 2

The correct option is C 1 lim_h→0In(1+2h) 2In(1+h)/h^2 =lim_h→0In(1+h)^2 In(1+2h)/h^2 =lim_h→0In ( (1+h)^2/1+2h )/h^2=lim_h→0In ( (1+h^2)/1+2h )/ ( h^2/1+2h ...

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

Byju's Answer

Standard XI

Mathematics

Standard Limits to Remove Indeterminate Form

The value of ...

Question

$The value of {lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} i s$

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

-1

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

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

Open in App

Solution

The correct option is B
-1

${lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} = {lim}_{h \to 0} \frac{I n (1 + h)^{2} - I n (1 + 2 h)}{h^{2}} = {lim}_{h \to 0} {\frac{I n (\frac{(1 + h)^{2}}{1 + 2 h})}{h}}^{2} = {lim}_{h \to 0} \frac{I n (\frac{(1 + h^{2})}{1 + 2 h})}{(\frac{h^{2}}{1 + 2 h}) (1 + 2 h)} = {lim}_{h \to 0} \frac{I n (1 + \frac{h^{2}}{1 + 2 h})}{(\frac{h^{2}}{.} 1 + 2 h)} \frac{1}{1 + 2 h} = - 1 [U s i n g {lim}_{x \to 0} \frac{l o g 1 + x}{x} = 1]$

Suggest Corrections

Similar questions

Q. The value of

lim h \to 0 \frac{ln (1 + 2 h) - 2 ln (1 + h)}{h^{2}}

The value of $p^{- 1} \cdot q \cdot q^{- 1} \cdot r \cdot r^{- 1} \cdot p$ is
(a) –1
(b) 0
(c) 1
(d) 2

lim h \to 0 \frac{- log (1 + 2 h) + 2 log (1 + h)}{h^{2}}

is equal to

Q. The value of tan 10° tan 15° tan 75° tan 80° is

(a) −1
(b) 0
(c) 1
(d) None of these

Q. The value of cos 1° cos 2° cos 3° ..... cos 180° is

(a) 1
(b) 0
(c) −1
(d) None of these

Join BYJU'S Learning Program

The value oflimh→0In(1+2h)−2In(1+h)h2is

The correct option is B -1 limh→0In(1+2h)−2In(1+h)h2=limh→0In(1+h)2−In(1+2h)h2=limh→0In((1+h)21+2h)h2=limh→0In((1+h2)1+2h)(h21+2h)(1+2h)=limh→0In(1+h21+2h)(h2.1+2h)11+2h=−1 [Usinglimx→0log1+xx=1]

$The value of {lim}_{h \to 0} \frac{I n (1 + 2 h) - 2 I n (1 + h)}{h^{2}} i s$