Limit's the Sky... (Problem $1$ )

Calculus Level 2

$\lim_{x \to 0} \left(\left(1 + \frac 1{x^2} \right)^x + \frac {2x}{\sin x + 1} - 1 \right) = \ ?$

1 solution

A Former Brilliant Member
Jul 29, 2020

$\lim_{x \to 0}\begin{cases} \left ( 1+\cfrac{1}{0^2}\right )^0=1\\ \cfrac{2\cdot0}{0+1}=0\\ \end{cases}$ $1+0-1=0$

Very good! Use the limit symbol, though!

Yajat Shamji - 10 months, 2 weeks ago

I haven't learned calculus yet, so I don't know the correct form of it :(

A Former Brilliant Member - 10 months, 2 weeks ago

$\lim_{x \to 0}$

Yajat Shamji - 10 months, 2 weeks ago

@Yajat Shamji – Yeah. I know the $\LaTeX$ code of lim, but where should I include it?

A Former Brilliant Member - 10 months, 2 weeks ago

@A Former Brilliant Member – Use it before your calculations.

Yajat Shamji - 10 months, 2 weeks ago