Did i solved this limit correctly

  • #1
1,395
0
[tex]
lim_{x->\infty}\frac{3x-cosx}{4x+sinx}=lim_{x->\infty}\frac{\frac{3x-cosx}{x}}{\frac{4x+sinx}{x}}=lim_{x->\infty}\frac{3-0}{4-0}=0.75
[/tex]

i thought bounded/infinity=0


??
 

Answers and Replies

  • #2
34,552
6,265
You have the correct limit, but you should remove the limit operator from the expression below, since you have already "passed to the limit" in the previous expression:
[tex]lim_{x->\infty}\frac{3-0}{4-0}=0.75 [/tex]
 
  • #3
1,395
0
thanks
 

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