A limit of a function problem by Huzaifa

Calculus Level 1

lim ⁡ x → 0 sin ⁡ ( 5 x ) x = ? \large \lim_{x\to0} \dfrac{\sin(5x)}x = \, ?

x x 5 5 5 x 5x x 2 x^2

Encantador Huzaifa by Encantador Huzaifa , 25, Pakistan

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3 solutions

As L i m i t Limit S i n Q x Q x \frac{SinQx}{Qx} =1

x→0

We need to multiply and divide by 5 so,

  • L i m i t Limit S i n 5 x x \frac{Sin5x}{x} x 5 5 \frac{5}{5}

  • L i m i t Limit S i n 5 x 5 x \frac{Sin5x}{5x} x 5

  • 1x5

  • 5
4 Helpful 0 Interesting 0 Brilliant 0 Confused
Stranger Rr
Sep 11, 2016

By L'Hospital rule, differentiate numerator and denominator individually by 'x'. Then apply limit to get 5.

i.e., Lim sin5x/x |x=0 = Lim 5cos5x/1 |x=0 = 5.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Rico Lee
Sep 10, 2016

Lol I don't even know how to do calculus and I guessed correctly. Well, I mean, I was like well if you plug in zero that can't work because denominator will be zero. Well, what if I just randomly cancelled out the x and yay I got 5.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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