limx-->2 ( sin(2x-4)/(x-2)

limx-->2 ( sin(2x-4)/(x-2)

limx--> 0+ (1/sqrtx) - ( 1/sqrtx^2+x) Please show me steps. Im really confused.

limx-->positive infty sqrt(3x^5-2x)-sqrt(3x^5-7x) I first rationalized the whole thing. After this , i divided the top and the bottom by x^5 but the answer I got is not the...

limx-->positive infty sqrt(9x^3+x) -x^3/2) The only one thats under the square root is 9x^3+x I dont know how to solve is

lim as x tends to positive infinity sqrt(3x^5-2x)-sqrt(3x^5-7x) Answer I got : (5)/(2√3)

lim x--> -infty ( 4x-3)/(25x^2+4x) Im not sure if this is a legititamte way of solving but heres how I attempted: I took the highest power at the bottom and top and...

Limx--->(pi/2) secx-tanx How do i solve this without hopital rule?

lim as x tends to positive infty (x+√x)/(x2+x+1)

I dont undestand why te answer to this is 1/sqrt2 limx-->0 ( sqrt2+x^2)-(sqrt2-x^2) /(x^2)

lim x--> 0+ [ (1/√x) - (1/sqrt(x2+x) ]

lim as x tends to positive infinity (x+√x)/(x2+x+1)

lim--> -infty (4+6e^2t)/(5-9e^3t) I really dont understand it

limx--> 5 ( | x-7 |) / (x-7) The right answer is -1 but I dont understand why. When I took the limit from both sides, they were different so I thought...

Hi, I really need help with the following : a) limx--->0 (1-cos2x)/(x) b) limx--> 0 (1-cos2t) /(sin^2(3t))

lim as x tends to -infinity x3/5 - x1/5 My first guess is to tuen the xs into a radical. After this im lost because the root will be 5 so what will happen then ? please...

limx-->positive infinity tan ( x/x^2+1) limx--> negative infinity cot ( pix/4x+1) This is so confusing for me, i would lice a detailed step...

limx--> -infty x3/5 - x1/5

limx-->infty (√3x5-2x) - (√3x5-7x) *In both brackets, the square root is over the entire polynomial.* I tries solving by first rationalizing, and then when it...

Find the limit as x approaches π (pi).

Lim as x tends to positive infinity ( √3x5-2x) - ( √3x5-7x) The square root is in the first and the second bracket is over the the whole polynomial. But both brackets...

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