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\[\lim_{x \rightarrow 0} \frac{ \sqrt[5]{1+2x}-1 }{ \sin x }\]

try l`hopital

sorry, I forgot to mention that we haven't learnt about l`hopital

nvm

you learned Taylors series?

Nope

expand the root term by binomial theorem

then there will be term like
10x/sinx
and
x/sinx=1
for given condition
so ans =10

while opening with binomial neglect higher degree terms

no see it i have shown|dw:1353844296253:dw|

and i have done mistake while doing this it is 2/5=0.4
so sorry for that

Could you explain how did you get
\[\sqrt[5]{1+2n}=1+\frac{ 2 }{ 5 }n + ....?\]

I can't use l'hoptial's rule to find the limit

i do not get you.Do you mean ,you are not allowed to use l hopitals rule or you can not evaluate.

We haven't studied about that rule yet

do u want to learn it?

@RajshikharGupta 's method seems to be the shortest to me,

are you asking how do we get (1+x)^n = 1+xn (given |x|<1 ) ?

Yeah, I don't underst that part

hmm,,heard of taylor series expansion ?

No

make that substitution in your question..
you'll reach the ans directly then..