Lim x - 0tan 2 3 x-x2

Lim_x→ 0tan^2 3x/x^2 = (lim_x→ 0tan 3x/x )^2= (lim_x→ 0tan 3x/3x )^2× 9 =1 × 9 [ ∵lim_x→ 0tanx/x =1 ] =9 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Sandwich Theorem

lim x → 0tan ...

Question

${lim}_{x \to 0} \frac{t a n^{2} 3 x}{x^{2}}$

Open in App

Solution

Suggest Corrections

Similar questions

lim x \to 0 \frac{{tan}^{2} 3 x}{x^{2}} =

lim x \to 0 \frac{{tan}^{2} 3 x}{x^{2}}

Evaluate the following one sided limits:

$(i) {lim}_{x \to 2^{+}} \frac{x - 3}{x^{2} - 4}$

$(i i) {lim}_{x \to 2^{-}} \frac{x - 3}{x^{2} - 4}$

$(i i i) {lim}_{x \to 0^{+}} \frac{1}{3 x}$

$(i v) {lim}_{x \to 8^{+}} \frac{2 x}{x + 8}$

$(v) {lim}_{x \to 0^{+}} \frac{2}{x^{\frac{1}{5}}}$

$(v i) {lim}_{x \to \frac{π^{-}}{2}} t a n x$

$(v i i) {lim}_{x \to \frac{π}{2^{+}}} s e c x$

$(v i i i) {lim}_{x \to 0^{-}} \frac{x^{2} - 3 x + 2}{x^{3} - 2 x^{2}}$

$(i x) {lim}_{x \to - 2^{+}} \frac{x^{2} - 1}{2 x + 4}$

$(x) {lim}_{x \to 0^{+}} (2 - c o t x)$

$(x i) {lim}_{x \to 0^{-}} 1 + c o s e c x$

\lim_{x \to \infty} (\sqrt{x^{2} + x + 1} - x) \neq \lim_{x \to \infty} (\sqrt{x^{2} + 1} - x)

Show that ${lim}_{x \to \infty} (\sqrt{x^{2} + x + 1 - x}) \neq {lim}_{x \to \infty} (\sqrt{x^{2} + 1} - x)$

Join BYJU'S Learning Program