Solve : calculate the limit

Question

Calculate the limit 

$$\lim_{x\rightarrow 1^- } \sqrt[3]{x+1}\: ln(x+1)$$

(A) $1$

(B) $0$

(C) $2$

(D) Does not exist

Comments

x tends to left side of 1 in limits ?? then you should give something like graph that explains behaviour when x approaches from left side of 1....

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is it D)...??

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Ans given B. I cant understand how .

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if i use general method ,h-->0 , and put x= 1-h then i got 0.69...

Answer 1

answer = none of these

$$\large \begin{align*} \lim_{x\rightarrow 1^-} \sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{x+1} \ln(x+1) &= \lim_{h\rightarrow 0} \sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{1-h+1} \ln(1-h+1) \\ &=\sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{2} \ln(2) \end{align*}$$

Comments

question meant -1?

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left hand limit ....rt?

and it give 0.69..@ amar

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yes, question asks its clearly, what if $\large \lim_{x\rightarrow 1^-}$

+ if it is to be intuitively be red as $\large \lim_{x\rightarrow -1}$ then after using L'Hospital rule on $\lim_{x\rightarrow -1} \frac{\ln(x+1)}{(x+1)^{-1/3}}$ it will evaluate to $0$

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yes, your answer is simple as compared to others...nice,,,

Answer 2

Question is misprinted....

It should be x-> -1.

Solve using x-> -1 ,apply L'hospital rule , u get ans '0'