আমাদের টেলিগ্রাম চ্যানেলে যুক্ত হোন

Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{xtanx}{1 – cosx}$

by

Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{xtanx}{1 – cosx}$

$\lim_{x\rightarrow 0}\frac{xtanx}{1 – cosx}$
= $\lim_{x\rightarrow 0}\frac{xsinx}{\cos x(1 – cos x)}$
=$\lim_{x\rightarrow 0}\frac{xsinx(1 + cos x)}{\cos x(1 – cos x)(1 + cos x)}$
= $\lim_{x\rightarrow 0}\frac{xsinx(1 + cos x)}{\cos x(1 – cos^2x)}$
= $\lim_{x\rightarrow 0}\frac{xsinx(1 + cos x)}{\cos xsin^2x}$
= $\lim_{x\rightarrow 0}\frac{x(1 + cos x)}{\cos xsinx}$
= $\lim_{x\rightarrow 0}\frac{2x(1 + cos x)}{2\cos xsinx}$
= $\lim_{x\rightarrow 0}\frac{(1 + cos x)}{\frac{\sin 2x}{2}}$
= $\frac{\lim_{x\rightarrow 0}(1 + cos x)}{\lim_{x\rightarrow 0}\frac{\sin 2x}{2}}$
= $\frac{1 + 1}{1}$
= 2

আরও দেখুন

  1. Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{1 – cosx}{x^2}$
  2. Limit x Tends to Zero $\lim_{x\rightarrow 0}\frac{1 – cosx}{\sin^2x}$
  3. Limit X Tends to Zero $\lim_{x\rightarrow 0}\frac{1 – cos4x}{x^2}$
  4. Limit h tends to zero $\lim_{h\rightarrow 0}\frac{1 – cosh}{hsinh}$
  5. Limit h Tends to 0 $\lim_{h\rightarrow 0}\frac{\cos ah – cosbh}{h^2}$

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