If limx→∞[x2+1x+1−ax−b]=0\lim_{x \to \infty} \left[ \frac{x^2 + 1}{x + 1} - ax - b \right] = 0x→∞lim[x+1x2+1−ax−b]=0, then (a,b) is
limn→∞1+2+3+…+nn2\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}n→∞limn21+2+3+…+n is equivalent to
What is the value of limx→12x3−1728x−12?\lim _{x \rightarrow 12} \frac{x^3-1728}{x-12} ?x→12limx−12x3−1728?
Compute
limx→4(x2−7x+12)x−4\lim _{x \rightarrow 4} \frac{\left(x^2-7 x+12\right)}{x-4}x→4limx−4(x2−7x+12)
Find limx→π4cot3x−tanxcos(x+π4).\text { Find } \lim _{x \rightarrow \frac{\pi}{4}} \frac{\cot ^3 x-\tan x}{\cos \left(x+\frac{\pi}{4}\right)}. Find x→4πlimcos(x+4π)cot3x−tanx.
If f(16)=16 and f′(16)=5, then limx→16f(x)−4x−4= ? \text { If } f(16)=16 \text { and } f^{\prime}(16)=5 \text {, then } \lim _{x \rightarrow 16} \frac{\sqrt{f(x)}-4}{\sqrt{x}-4}=\text { ? } If f(16)=16 and f′(16)=5, then x→16limx−4f(x)−4= ?
The value of limx→∞(2x−12x+3)x+12 is: \text { The value of } \lim _{x \rightarrow \infty}\left(\frac{2 x-1}{2 x+3}\right)^{\frac{x+1}{2}} \text { is: } The value of x→∞lim(2x+32x−1)2x+1 is:
limθ→01−cosmθ1−cosnθ= ?\lim_{\theta \to 0} \frac{1 - \cos{m\theta}}{1 - \cos{n\theta}} = \, ?θ→0lim1−cosnθ1−cosmθ=?
Suggested Test Series